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IN  MEMORIAM 
FLOR1AN  CAJORI 


Ia&w 


ELEMENTARY   ALGEBRA 


BY 


FREDERICK   H.   SOMERVILLE,   B.S. 

THE  WILLIAM  PENN  CHARTER  SCHOOL,   PHILADELPHIA 


NEW  YORK  .^CINCINNATI-:.  CHICAGO 

AMERICAN    BOOK    COMPANY 


Copyright,  1908,  by 
FREDERICK   H.   SOMERVILLE. 

Entered  at  Stationers'  Hall,  London. 

el.  alo. 
W.  P.     I 


\JMf 


PREFACE 

The  plan  of  the  early  pages  of  this  text  is  to  offer  a  gradual 
introduction  to  the  subject  without  plunging  the  young  student 
too  deeply  into  new  difficulties.  The  treatment  of  negative 
numbers  as  a  natural  extension  of  the  familiar  arithmetical 
numbers,  and  a  generous  amount  of  detail  in  the  explanations 
of  the  early  chapters,  serve  to  clarify  the  beginnings  of  a  sub- 
ject so  often  troublesome.  New  processes  are  accomplished  by 
the  use  of  the  simplest  of  symbols,  and  the  undivided  atten- 
tion of  the  student  is  centered  upon  the  new  elements  of  the 
operation  at  hand.  Definitions  are  introduced  only  as  rapidly 
as  new  processes  call  for  them,  and  the  young  mind  is  not 
confused  and  discouraged  at  the  outset  by  the  attempt  to  learn 
the  meaning  of  a  bewildering  mass  of  strange  terms. 

In  arrangement  the  book  does  not  differ  widely  from  the 
general  scheme  of  the  standard  texts,  but  in  some  details  there 
are  changes  that  have  been  found  to  be  of  genuine  value  in  the 
classroom.  For  example,  in  Factoring,  the  simple  and  the 
difficult  types  are  separated ;  an  elementary  course  gives  a 
treatment  free  from  complex  forms,  while  a  supplementary 
section  provides  for  the  preparation  of  college  requirements. 
The  Lowest  Common  Multiple  and  its  immediate  application 
to  Addition  of  Fractions  form  a  single  chapter,  and  the  plan 
suggested  by  this  order  has  proved  to  be  natural,  practical,  and 
sound.  The  classification  of  Simple  Simultaneous  Equations 
and  of  the  Theory  of  Exponents  is  both  new  and  teachable, 
and  the  logical  arrangement  of  Affected  Quadratic  Equations 
provides  a  chapter  that,  while  omitting  no  essential,  gives  a 
brief  and  clear  general  treatment  of  this  important  subject. 

Throughout  the  early  chapters  exercises  for  oral  drill  are 
frequent,  and  their  introduction  confines  the  simplest  types 

3 


rz*yf\£yc\*Jtz 


4  PREFACE 

of  the  problems  to  the  oral  discussion  of  the  classroom.  Such 
discussions  are  of  great  value  in  a  live  class,  and  if  supple- 
mented by  a  practice  of  reading  by  members  of  the  class  each 
step  of  every  new  illustrative  solution,  a  most  gratifying 
progress  results.  The  written  exercises  consist  of  new  prob- 
lems carefully  graded,  and  the  frequent  reviews  are  constructed 
on  the  lines  of  recent  entrance  questions  of  the  leading  col- 
leges and  universities.  The  treatment  of  Graphs  is  full  and 
complete,  but  is  free  from  those  elements  of  the  advanced 
discussions  that  so  frequently  confuse  the  young  student.  A 
comprehensive  introduction  of  the  common  Physical  Formulas 
familiarizes  the  student  with  a  practical  branch  of  applied 
algebra;  and  this  is  accomplished  without  assuming  that  the 
teacher  is  an  expert  in  the  laboratory  practice  of  that  science. 

In  those  subjects  where  several  methods  of  procedure  are 
possible,  the  text  offers,  as  a  rule,  but  one.  To  select  arbi- 
trarily and  to  suggest  one  method  as  the  best  of  many  is  a 
matter  of  personal  choice  and  opinion,  and  the  only  claim  for 
the  single  methods  chosen  in  the  following  chapters  is  that 
they  are  uniformly  practical.  The  text  is  planned  on  the 
theory  that  one  practical  method  thoroughly  mastered  is 
sufficient  for  the  needs  of  the  young  student,  and  that  the 
elementary  classroom  has  neither  time  nor  need  for  those 
comparative  discussions  that  interest  only  the  mature  mind. 

The  author  gratefully  acknowledges  his  indebtedness  to 
those  friends  whose  suggestions  and  encouragement  have 
been  of  material  aid  in  the  preparation  of  this  text. 

FREDERICK  H.   SOMERVILLE. 

The  William  Penn  Charter  School, 
Philadelphia  . 


CONTENTS 

CHAPTER  PAGE 

I.    Introduction,  Symbols,  Negative  Numbers          ...  9 

Symbols  of  Operation 11 

Symbols  of  Quality 13 

Negative  Numbers 14 

Algebraic  Expressions 20 

II.    Addition,  Parentheses 22 

Addition  of  Monomials 22 

Addition  of  Polynomials 24 

Parentheses 26 

III.  Subtraction,  Review 32 

Subtraction  of  Polynomials       .         .        ....  34 

General  Review 35 

IV.  Multiplication        .        .        .        . 38 

Multiplication  of  a  Monomial  by  a  Monomial  ...  42 

Multiplication  of  a  Polynomial  by  a  Monomial         .        .  43 

Multiplication  of  a  Polynomial  by  a  Polynomial       .         .  45 

Multiplication  of  Miscellaneous  Types      ....  47 

V.    Division,  Review 52 

Division  of  a  Monomial  by  a  Monomial    ....  54 

Division  of  a  Polynomial  by  a  Monomial ....  55 

Division  of  a  Polynomial  by  a  Polynomial        .         .         .  ,56 

General  Review^ 62 

VI.    The  Linear  Equation,  The  Problem    .  .  .64 

General  Solution  of  the  Linear  Equation  ....  68 

The  Solution  of  Problems 70 

VII.     Substitution 84 

The  Use  of  Formulas 87 

VIII.     Special  Cases  in  Multiplication  and  Division     ...  89 

Multiplication 89 

Division 95 

5 


CONTENTS 


IX.    Factoring 

Expressions  having  the  Same  Monomial  Factor 

Trinomial  Expressions 

Binomial  Expressions 

Expressions  of  Four  or  More  Terms  factored  by  Grouping 
Supplementary  Factoring 


X.    Highest  Common  Factor 

Highest  Common  Factor  of  Monomials    . 

Highest  Common  Factor  of  Polynomials  by  Factoring 


XI.  Fractions,  Transformations  . 
The  Signs  of  a  Fraction  . 
Transformations  of  Fractions 


XII.     Fractions  (Continued),  Lowest  Common  Multiple,  Lowest 
Common  Denominator,  Addition  of  Fractions 
Lowest  Common  Multiple 
Lowest  Common  Multiple  of  Monomials  . 
Lowest  Common  Multiple  of  Polynomials  by  Factoring 
Lowest  Common  Denominator  .        . 

Addition  and  Subtraction  of  Fractions     . 

XIII.    Fractions  (Continued),  Multiplication,  Division,  The  Com 

plex  Form 

Multiplication  of  Fractions       ..... 

Division  of  Fractions 

The  Complex  Form 


XIV.     Fractional  and  Literal  Linear  Equations,  Problems 
Special  Forms  .        .        .        .      * . 
Problems  leading  to  Fractional  Linear  Equations 

XV.    Applications  of  General  Symbols,  Review 
The  General  Statement  of  a  Problem 
Use  of  Physical  Formulas 
Transformation  of  Formulas     . 
General  Review 

XVI.     Simultaneous  Linear  Equations,  Problems 
Elimination  by  Substitution 


CONTENTS 


CHAPTER  PAGE 

Elimination  by  Comparison 176 

Elimination  by  Addition  or  Subtraction  ....  176 

Systems  involving  Three  or  More  Unknown  Quantities    .  178 

Fractional  Forms 181 

Literal  Forms   . 185 

Problems  producing  Simultaneous  Linear  Equations        .  187 

The  Discussion  of  a  Problem 193 

XVII.     Graphical  Representation  of  Linear  Equations  .        .        .197 

Graph  of  a  Point 197 

Graph  of  a  Linear  Equation  in  Two  Unknown  Numbers  .  200 

Graphs  of  Simultaneous  Linear  Equations        .        .        .  202 

XVIII.    Involution  and  Evolution 206 

Involution 205 

Evolution 209 

Square  Root  of  Polynomials 213 

Square  Root  of  Arithmetical  Numbers     .        .        .        .217 

XIX.    Theory  of  Exponents 221 

The  Zero  Exponent 221 

The  Negative  Exponent    .        .        .        .                 .         .  222 

The  Fractional  Form  of  the  Exponent      ....  223 

Applications 227 

XX.     Radicals,  Imaginary  Numbers,  Review      ....  236 

Transformation  of  Radicals 237 

Operations  with  Radicals 242 

Equations  involving  Radical  Expressions    .     • .        .        .  252 

Imaginary  and  Complex  Numbers 254 

Operations  with  Imaginary  Numbers        ....  255 

General  Review         .        . 261 

XXI.     Quadratic  Equations 266 

Pure  Quadratic  Equations 266 

Affected  Quadratic  Equations 268 

Discussion  of  Affected  Quadratic  Equations     .         .         .277 

Graphs  of  Affected  Quadratic  Equations  ....  282 

XXII.     The  Quadratic  Form,.  Higher  Equations,  Irrational  Equa- 
tions        286 

The  Quadratic  Form '  .        .286 


CONTENTS 


CHAPTER  PAGE 

Higher  Equations  solved  by  Quadratic  Methods   .        .  286 

Irrational  Equations 292 

Physical  Formulas  involving  Quadratic  Equations         .  293 

XXIII.  Simultaneous  Quadratic  Equations,  Problems                 .  294 

Solution  by  Substitution 294 

Solution  by  Comparison  and  Factoring         .        .        .  295 

Solution  of  Symmetrical  Types      .        .                 .        .  296 

Solution  of  Miscellaneous  Types 298 

Graphs  of  Quadratic  Equations  in  Two  Variables          .  302 

Problems  producing  Quadratic  Equations     .         .        .  310 

XXIV.  Ratio,  Proportion,  Variation 315 

Katio 315 

Proportion     .                                  317 

Variation 330 

XXV.     The  Progressions 338 

Arithmetical  Progression 338 

Geometrical  Progression 348 

XXVI.     The  Binomial  Theorem  —  Positive  Integral  Exponent    .  358 

Proof  of  the  Binomial  Formula 359 

Applications 361 

XXVII.    Logarithms 364 

The  Parts  of  a  Logarithm 365 

The  Use  of  the  Four-place  Table 367 

The  Properties  of  Logarithms 374 

The  Cologarithm 376 

Use  of  Logarithms  in  Computations      ....  377 

Miscellaneous  Applications  of  Logarithms    .        .        .  379 

XXVIII.     Supplementary  Topics,  Review       .        .        .        .        .385 

The  Remainder  Theorem 385 

The  Factor  Theorem 385 

The  Theory  of  Divisors  of  Binomials    .        .        .        .  387 

H.  C.  F.  of  Expressions  not  readily  Factorable     .        .  389 

L.  C.  M.  of  Expressions  not  readily  Factorable     .        .  392 

Cube  Root  of  Polynomials 394 

Cube  Root  of  Arithmetical  Numbers     ....  396 

General  Review 398 

INDEX                                                                                                     .  405 


ELEMENTARY  ALGEBRA 

CHAPTER   I 
INTRODUCTION.     SYMBOLS.     NEGATIVE   NUMBERS 

1.  The  definite  number  symbols  of  arithmetic,  1, 2, 3, 4, 5,  etc., 
are  symbols  that  express  in  each  case  a  number  with  one 
definite  value. 

Thus,  3  units,  5  units,  7  units,  etc.,  represent  particular  groups,  the 
symbols,  3,  5,  and  7  having  each  a  particular  name  and  value  in  the 
number  system  that  we  have  learned  to  use. 

2.  The  general  number  symbols  of  algebra  are  symbols  that 
may  represent  not  one  alone  but  many  values,  and  for  these 
general  symbols  the  letters  of  the  alphabet  are  in  common  use. 

THE  ADVANTAGE  OF  THE  GENERAL  NUMBER  SYMBOL 

3.  Many  of  the  familiar  principles  of  arithmetic  may  be 
stated  much  more  briefly,  and  usually  with  greater  clearness, 
if  general  or  literal  number  symbols  are  employed.  To 
illustrate : 

(a)  The  area  of  a  rectangle  is  equal  to  the  product  of  its  height,  or 
altitude,  by  its  length,  or  base.  Or,  arith- 
metically, 

Area  =  altitude  x  base. 

**        Using  only  the  first  letters  of  each  word, 
we  may  write, 


Area  =  a  x  6, 


b 

and  this  latter  expression,  while  equally  clear 
in  meaning,  serves  as  a  general  expression  for  obtaining  the  area  of  any 
rectangle  with  any  values  of  a  and  b. 

9 


10       INTRODUCTION.     SYMBOLS.     NEGATIVE   NUMBERS 

(6)  The  familiar  problem  of  simple  interest  gives  arithmetically, 
Interest  =  principal  x  rate  x  time.    . 

A  much  simpler  form  of  expressing  the  same  principle  is  obtained 
here  as  above,  using  only  the  first  letters  of  each  element  for  the  general 
expression. 

Thus :  I  =  p  x  r  x  t. 

(c)  Two  principles  relating  to  the  operation  of  division  in  arithmetic 
may  be  clearly  and  briefly  expressed  by  the  use  of  literal  symbols. 

Thus  :     Dividend  -f-  divisor  =  quotient.  That  is,      —  =  q. 

d 

divisor  x  quotient  =  Dividend.  Or,        d  x  q  =  D. 

4.  In  a  more  extended  manner  the  literal  symbol  permits 
a  breadth  and  power  of  expression  not  hitherto  possible  with 
the  number  symbols  of  arithmetic.  Many  problems  involving 
unknown  quantities  are  readily  stated  and  solved  by  means  of 
general  symbols,  and  clearness  of  expression  in  such  problems 
is  invariably  gained  by  their  use.  To  illustrate  a  common  use 
of  literal  symbols,  consider  the  following  problem  : 

Two  brothers,  John  and  William,  possess  together  200 
books,  and  John  has  20  books  more  than  William.  Write 
expressions  that  clearly  state  these  conditions.  (Compare 
carefully  the  two  methods  of  expression.) 

The  Arithmetical  Expression 
(a)  The  number  of  John's  books  +  the  number  of  William's  books =200 
(6)  The  number  of  John's  books  —  the  number  of  William's  books  =  20 

The  Algebraic  Expression 
Let  us  assume  that  x  =  the  number  of  books  that  John  has, 
and  that  y  =  the  number  of  books  that  William  has. 

Then  from  the  conditions  given  in  the  problem  : 
(a)  x  +  y  =  200. 

(6)  x-y  =   20. 


THE  SYMBOLS  OF  OPERATION  11 

By  a  simple  process  the  values  of  x  and  y  are  readily  determined. 
From  this  parallel  between  arithmetical  and  algebraic  forms  of  ex- 
pression the  brevity  and  the  advantage  of  the  literal  or  general  symbol, 
for  number  is  clearly  manifest.  The  later  processes  of  algebra  will  con- 
stantly furnish  the  means  wherewith  we  may  broaden  our  power  of 
expression,  and  the  meaning  of  algebra  will  be  interpreted  as  merely 
an  extension  of  our  processes  with  number. 


THE   SYMBOLS  OF  OPERATION 

5.  The  principal  signs  for  operations  in  algebra  are  identical 
with  those  of  the  corresponding  operations  in  arithmetic. 

6.  Addition  is  indicated  by  the  "  pins  "  sign,  + . 

Thus,  a  +  b  is  the  indicated  sum  of  the  quantity  a  and  the  quantity  b. 
The  expression  is  read  "  a  plus  6." 

7.  Subtraction  is  indicated  by  the  "  minus  "  sign,  — . 

Thus,  a  —  b  is  the  indicated  difference  between  the  quantity  b  and  the 
quantity  a.     The  expression  is  read  "  a  minus  6." 

8.  Multiplication  is  usually  indicated  by  an  absence  of  sign 
between  the  quantities  to  be  multiplied. 

Thus,  ab  is  the  indicated  product  of  the  quantities  a  and  b. 

abx  is  the  indicated  product  of  the  quantities  a,  6,  and  x. 

Sometimes  a  dot  is  used  to  indicate  a  multiplication. 
Thus,  a  •  b  is  the  product  of  a  and  b. 

The  ordinary  symbol,  "  x ,"  is  occasionally  used  in  algebraic 
expression. 

An  indicated  product  may  be  read  by  the  use  of  the  word 
"  times  "  or  by  reading  the  literal  symbols  only. 

Thus,  ab  may  be  read  "a  times  6,"  or  simply  " a&." 

9.  Division  is  indicated  by  the  sign  "  -j-,"  or  by  writing  in 
the  fractional  form. 


12       INTRODUCTION.    SYMBOLS.     NEGATIVE  NUMBERS 

Thus,  a  -*-  b  is  the  indicated  quotient  of  the  quantity  a  divided  by  the 
quantity  b. 

-  is  the  fractional  form  for  the  same  indicated  quotient. 
b 

Both  forms  are  read  u  a  divided  by  6." 

10.  Indicated  operations  are  of  constant  occurrence  in  alge- 
braic processes,  for  the  literal  symbols  do  not  permit  the 
combining  of  two  or  more  into  a  single  symbol  as  in  the  case 
of  numerals.     Thus: 

Arithmetically,  5  +  3  +  7  may  be  written  "  15,"  for  the  symbol  15  is 
the  symbol  for  the  group  made  up  of  the  three  groups,  5,  3,  and  7. 

Algebraically,  a  +  b  +  c  cannot  be  rewritten  unless  particular  values 
are  assigned  to  the  symbols  a,  &,  and  c.     The  sum  is  an  indicated  result. 

Algebraic  expression,  therefore,  confines  us  to  a  constant  use 
of  indicated  operations,  and  we  must  clearly  understand  the 
meaning  of : 

I.    An  Indicated  Addition,  a  +-  b. 

II.    An  Indicated  Subtraction,  a  —  b. 

III.  An  Indicated  Multiplication,     ab. 

IV.  An  Indicated  Division,  -• 

b 

11.  Equality  of  quantities  or  expressions  in  algebra  is  indi- 
cated by  the  sign  of  equality,  =,  read  "equals,"  or  "is  equal  to." 

Thus,  a  +  6  =  c  +  d 

is  an  indicated  equality  between  two  quantities,  a  +  b  and  c  +  d. 

12.  If  two  or  more  numbers  are  multiplied  together,  each  of 
them,  or  the  product  of  two  or  more  of  them,  is  a  factor  of  the 
product ;  and  any  factor  of  a  product  may  be  considered  the 
coefficient  of  the  product  of  the  other  factors.     Thus : 

In  5  a,  5  is  the  coefficient  of  a.  In  acmx,  a  is  the  coefficient  of  cmx, 

In  ax,  a  is  the  coefficient   of  x,  or  ac  may  be  the  coefficient  of 

or  x  is  the  coefficient  of  a.  mx,  etc. 


THE   SYMBOLS   OF   QUALITY  13 

Coefficients  are  the  direct  results  of  additions,  for 

5  a  is  merely  an  abbreviation  of  a  +  a  +  a  +  a  + a. 
4  xy  is  an  abbreviation  of  xy  +  xy  +  xy  +  xy. 

If  the  coefficient  of  a  quantity  is  "  unity  "  or  "  1,"  it  is  not 
usually  written  or  read. 

Thus,        a  is  the  same  as  1  a.  xy  is  the  same  as  1  xy. 

13.  The  parenthesis  is  used  to  indicate  that  two  or  more 
quantities  are  to  be  treated  as  a  single  quantity.  The  ordinary 
form,  (  ),  is  most  common.  For  clearness  in  the  discussion  of 
elementary  principles,  the  parenthesis  will  frequently  be  made 
use  of  to  inclose  single  quantities. 

14.  An  axiom  is  a  statement  of  a  truth  so  simple  as  to  be 
accepted  without  proof.  Two  of  the  axioms  necessary  in 
early  discussions  are : 

Axiom  1.     If  equals  are  added  to  equals,  the  sums  are  equal. 

Axiom  2.  If  equals  are  subtracted  from  equals,  the  re- 
mainders are  equal. 

THE  SYMBOLS  OF  QUALITY 

15.  In  scientific  and  in  many  everyday  discussions  greater 
clearness  and  convenience  have  resulted  from  a  definite  method 
of  indicating  opposition  of  quality.     For  example : 

Temperature  above  and  below  the  zero  point, 

Latitude  north  and  south  of  the  equator, 

Assets,  or  possessions,  and  liabilities,  or  debts,  in  business,  etc., 

represent  cases  in  which  direct  opposites  of  quality  or  kind 
are  under  discussion;  hence,  a  need  exists  for  a  form  of  ex- 
pression that  shall  indicate  kind  as  well  as  amount,  quality  as 
well  as  quantity. 

To  supply  this  need  the  plus  and  minus  signs  are  in  general 
use,  their  direct  opposition  making  them  useful  as  signs  of 


14       INTRODUCTION.     SYMBOLS.    NEGATIVE   NUMBERS 

quality  as  well  as  of   operation;    and  we  will  now   consider 
their  use  as 

THE  +  AND  -  SIGNS  OP  QUALITY 

16.  A  few  selected  examples  of  common  occurrence  clearly 
illustrate  the  application  of  quality  signs  to  opposites  in  kind. 

In  the  laboratory :  % 

Temperature  above  0°  is  considered  as  +. 
Temperature  below  0°  is  considered  as  — . 

In  navigation : 

Latitude  north  of  the  equator  is  considered  as  +. 
Latitude  south  of  the  equator  is  considered  as  — . 

In  business  administration : 

Assets,  or  possessions,  are  considered  as  +  . 
Liabilities,  or  debts,  are  considered  as  — . 

The  following  illustration  emphasizes  the  advantage  of  in- 
dicating opposites  in  kind  by  the  use  of  the  -J-  and  —  signs 
of  quality. 

A  thermometer  registers  10°  above  0  at  8  a.m.,  0 

15°  above  0  at  11  a.m.,  5°  below  0  at  4  p.m.,  and  '  •■  '         ,M 

1am        -4-  1  o 

10°  below  at  10  p.m.     At  the  right  we  have  tabu-  0 

lated  the  conditions  in  a  concise  form  made  possible         .,_  ,  _0 

i     ^         u  .1.  £         ,.4      •  10  p.m.      —  10° 

only  through  the  use  of  quality  signs. 

By  applying  this  idea  of  opposition  to  arithmetical  numbers 
we  may  establish 

NEGATIVE  NUMBERS 

17.  It  is  first  necessary  to  show  that  there  exists  a  need  for 
a  definite  method  of  indicating  opposition  of  quality  in  number. 

Consider  the  subtractions : 

(1)  5-4=  (2)  5-5=  (3)  5-6  = 

There  are  three  definite  cases  included. 

In  (1)  a  subtrahend  less  than  the  minuend. 

In  (2)  a  subtrahend  equal  to  the  minuend. 

In  (3)  a  subtrahend  greater  than  the  minuend. 


NEGATIVE  NUMBERS  15 

The  first  two  cases  are  familiar  in  arithmetical  processes,  but 
the  third  raises  a  new  question,  and  we  ask, 
5  —  6  =  what  number  ? 

18.  On  any  convenient  straight  line  denote  the  middle 
point  by  "  0,"  and  mark  off  equal  points  of  division  both  to 
the  right  and  to  the  left  of  0. 

The  two  directions  from  0  are  dearly  defined  cases  of  opposition, 
and  this/  opposition  may  be  indicated  by  marking  the  suc- 
cessive division  points  at  the  right  of  0  with  numerals  having 
plus  signs, 

-5      -4      -3      -2       -1         O      +1      +2      +3      +4      +5 

1 

and,  similarly,  the  division  points  left  of  0  with  numerals  hav- 
ing minus  signs. 

The  result  is  a  series  of  positive  and  negative  numbers 
established  from  a  given  point  which  we   may  call   "  zero." 

With  this  extended  number  system  we  may  at  once  obtain  a 
clear  and  logical  answer  for  the  question  raised  above.  The 
minuend  remaining  the  same  in  each  case,  the  result  for  each 
subtraction  is  established  by  merely  counting  off  the  subtrahend 
from  the  minuend,  the  direction  of  counting  being  toward  zero. 
Therefore, 

For  (1)  The  subtraction  of  a  positive  number  from  a  greater 
positive  number  gives  a  positive  result. 

Or,  5-4  =  1. 

For  (2)  The  subtraction  of  a  positive  number  from  an  equal 
positive  number  gives  a  zero  result. 

Or,  5-5  =  0. 

For  (3)  The  subtraction  of  a  positive  number  from  a  less  posi- 
tive number  gives  a  negative  result. 

Or,  5  -  6  =  -  1. 


16       INTRODUCTION.     SYMBOLS.    NEGATIVE  NUMBERS 
We  may  conclude,  therefore,  that : 

19.  The  need  of  a  negative  number  system  is  that  subtraction 
may  be  always  possible.  On  the  principle  of  opposition  the  idea 
of  negative  number  is  as  firmly  established  as  that  of  positive 
number,  and  the  definition  of  algebra  as  an  extension  of  number 
is  even  further  warranted. 

20.  The  extension  of  arithmetical  number  to  include  nega- 
tive as  well  as  positive  number  establishes  algebraic  number. 

21.  It  is  often  necessary  to  refer  to  the  magnitude  of  a  num- 
ber regardless  of  its  quality,  the  number  of  units  in  the  group 
being  the  only  consideration. 

Thus,  -f  5  and  —  5  have  the  same  magnitude,  or  absolute  value,  for 
each  stands  for  the  same  group  idea  in  the  numeral  classification.  They 
differ  in  their  qualities,  however,  being  exact  opposites.  In  like  manner, 
-f  a  and  —  a,  +  xy  and  —  xy,  are  of  the  same  absolute  value  in  each 
respective  case,  no  matter  what  values  are  represented  by  a  or  by  xy. 

POSITIVE  AND  NEGATIVE  NUMBERS  COMBINED 

The  simplest  combination  of  algebraic  numbers  is  addition, 
and  since  two  groups  of  opposite  kinds  result  from  the  positive 
and  negative  qualities,  we  must  consider  an  elementary  dis- 
cussion of  algebraic  addition  under  three  heads. 

I.  Positive  Units  +  Positive  Units. 

If  a  rise  in  temperature  is  16°  -f  15° 

and  a  further  rise  of  10°  occurs,  4-  10° 

we  have  a  final  reading  of  .        .        .  .        .      +  25° 

II.  Negative  Units  +  Negative  Units 

If  a  fall  in  temperature  is  12°  — 12° 

and  a  further  fall  of  16°  occurs,  —  16° 


we  have  a  final  reading  of —  28c 


POSITIVE  AND  NEGATIVE   NUMBERS  COMBINED       17 

III.   Positive  Units  +  Negative  Units. 

If  a  rise  in  temperature  is  20°  and  -f  20° 

an  immediate  fall  of  15°  occurs,  —  15° 

we  have  finally +    5° 

If  a  fall  in  temperature  is  30°  and  _  30° 

an  immediate  rise  of  24°  occurs,  +  24° 


we  have  finally —    6° 

The  student  should  verify  these  illustrations  by  making  a 

sketch  of  a  thermometer  and  applying  each  case  given  above. 

In  general,  addition  of  algebraic  numbers  results  as  follows  : 


(1)  If  a  and  b  are  positive  numbers : 

(+«)  +  (+&)  =  +«  +  &.  0 i     .     .     . 

Numerical  Illustration :  V        *5      ,  p       "f"3    J 

(+5)  +  (+3)  =  +  5  +  3=+8. 


(2)  If  a  and  b  are  negative  numbers : 

(-«)  +  (- 6)  =  -a-  b  ,    ^     ,      ,      ,    1     ,      ,    O 

Numerical  Illustration :  V^3 Q       5  J 

(_5)  +  (-3)  =  -5-3=  -8. 

(3)  If  a  is  positive  and  b  negative  : 
(+«)  +  (-&)=  +a-b.  0  +5.      .      . 


Numerical  Illustration :  ^-+2 

(+5)  +  (-3)=+5-3=+2. 

(4)  If  a  is  negative  and  b  positive  : 

(-a)  +  (+6)  =  -a  +  6.  -5  Q 

Numerical  Illustration:  r     +3      ' — 2— '" 

(_5)  +  (+3)=-5  +  3  =  -2. 

From  the  four  cases  we  may  state  the  general  principles  for 
combining  by  addition  any  given  groups  of  positive  quantities, 
negative  quantities,  or  positive  and  negative  quantities : 

SOM.    EL.    ALG.  —  2 


18       INTRODUCTION.     SYMBOLS.     NEGATIVE  NUMBERS 

22.  The  sum  of  two  groups  of  plus,  or  positive,  units  is  a  posi- 
tive quantity. 

23.  TJie  sum  of  two  groups  of  minus,  or  negative,  units  is  a 
negative  quantity. 

24.  The  sum  of  two  groups  of  units  of  opposite  quality  is  posi- 
tive if  the  number  of  units  in  the  positive  group  is  the  greater,  but 
negative  if  the  number  of  units  in  the  negative  group  is  the 
greater.  * 

JTrom  Art.  24  we  derive  the  following  important  principle : 

25.  The  sum  of  two  units  of  the  same  absolute  value  but  oppo- 
site in  sign  is  0. 

In  general:  (+«)  +  (-«)  =  + a- a  =  0. 

Numerical  Illustration :  (+5) +  (-5)  =  +5-5  =  0. 

Exercise  1 

1.  If  distances  to  the  right  are  to  be  considered  as  posi- 
tive in  a  discussion,  what  shall  we  consider  distances  to  the 
left? 

2.  If  the  year  20  a.d.  is  considered  as  the  year  "  -f  20," 
how  shall  we  express  the  year  20  b.c.  ? 

3.  Draw  a  sketch  of  a  thermometer,  and  indicate  upon  it 
the  temperature  points  +  35°,  — 18°,  + 12°,  and  — 12°. 

4.  On  your  sketch  determine  the  number  of  degrees  passed 
through  if  the  temperature  rises  from  8°  to  35° ;  *  from  0  to 
15°;   from  -  15°  to  10°. 

5.  On  your  sketch  determine  the  number  of  degrees  passed 
through  in  a  fall  of  temperature  from  35°  to  10° ;  from  20°  to 
-10°. 

*  Note  that  the  quality  of  a  unit  is  considered  plus  when  no  sign  of 
quality  is  given. 


POSITIVE   AND   NEGATIVE   NUMBERS   COMBINED        19 

6.  Show  by  your  sketch  that  a  rise  of  25°  followed  by  a 
fall  of  15°  results  in  an  actual  change  of  10°  from  the  starting 
point. 

7.  In  example  6  is  the  conditi6n  true  for  the  single  case 
when  you  begin  at  0°,  or  is  the  result  the  same  no  matter  at 
what  point  you  begin  ?     Illustrate. 

8.  Show  that  a  fall  of  18°  from  the  zero  point  succeeded  by 
a  rise  of  30°  results  in  a  final  reading  of  + 12°. 

9.  Show  that  a  rise  of  10°  from  the  zero  point  succeeded  by 
a  further  rise  of  18°,  and  later  by  a  fall  of  40°,  results  in  a 
final  reading  of  12°  below  zero.  Express  "  12°  below  zero  "  in 
a  simpler  form. 

10.  From  your  sketch  determine  the  final  reading  when, 
after  a  rise  of  40°  from  0,  there  occurs  a  fall  of  15°,  a  succeed- 
ing rise  of  7°,  and  another  fall  of  35°.  Express  these  changes 
with  proper  signs. 

11.  C  is  a  point  on  the  line  AC  B 
AB.     A  traveler  starts  at  C,  goes                         +10        ^ 

10  miles  toward  B,  turns  back  7  <    ~+9 

miles    toward    A,   and    returns    9 

miles  toward  B.  Determine  his  final  distance  from  G,  and 
also  his  position  at  either  the  right  or  the  left  of  C.  (Assume 
distances  to  the  right  of  C  as  +.)  Would  the  information 
given  be  sufficient  to  determine  the  result  without  using  the 
sketch  ? 

12.  Determine  the  result  of  a  journey  8  miles  from  C  toward 
B,  returning  6  miles  toward  A.  Give  the  total  distance  trav- 
eled and  the  final  position. 

13.  Determine  the  result  of  a  journey  9  miles  from  C 
toward  B,  16  miles  back  toward  A,  and  then  8  miles  toward  B. 
Make  a  drawing  similar  to  that  illustrating  example  11,  and 
prove  your  answer  by  reference  to  the  drawing. 


20       INTRODUCTION.     SYMBOLS.     NEGATIVE   NUMBERS 

14.  Determine  the  result  of  a  journey  17  miles  toward  B, 
10  miles  back  toward  A,  3  miles  more  toward  A,  back  19  miles 
toward  B,  the  starting  point  being  at  G.  Make  a  drawing 
illustrating  the  entire  journey. 

15.  With  the  same  distances  and  directions  as  in  problem  14, 
determine  the  result  if  the  starting  point  had  been  at  A,  giv- 
ing for  the  answer  the  final  distance  of  the  traveler  from  G  as 
well  as  from  A.     (Assume  that  from  G  to  A  is  30  miles.) 


ALGEBRAIC  EXPRESSIONS 

26.  An  algebraic  expression  is  an  algebraic  symbol,  or  group 
of  algebraic  symbols,  representing  some  quantity.  An  ex- 
pression is  numerical  when  made  up  wholly  of  numerical 
symbols,  and  is  literal  when  made  up  wholly  or  in  part  of 
literal  symbols. 

20  +  10  —  13  x  2  is  a  numerical  expression. 

ab  +  mn  —  xy  and  5  a  —  6  ex  —  12  are  literal  expressions. 

27.  The  parts  of  an  expression  connected  by  the  +  or  - 
signs  are  the  terms. 

In  the  expression 

ab  +  (C  -  d)  _  mn  +  «  +  »  _  g .4 -  « » -P 
b  -y       2(d  +  k) 
the  terms  are 

ab,  +  (c  —  d),  —  mn,  +  a  ~  x,  and 


b  -  y  2(d  +  k) 

A  parenthesis,  or  a  sign  of  the  same  significance,  may  inclose  a  group 
as  one  term.     A  fraction  as  an  indicated  quotient  is  also  a  single  term. 

The  sign  between  two  terms  is  the  sign  of  the  term  following. 

The  sign  of  the  first  term  of  an  expression  is  not  usually  written  if  it 
is  +.  In  the  term  ab  of  the  given  expression  both  the  sign,  +,  and  the 
coefficient,  1,  are  understood. 


ALGEBRAIC  EXPRESSIONS  21 

28.  Terms  not  differing  excepting  as  to  their  numerical 
coefficients  are  like  or  similar  terms. 

5 ax  and  —  Sax  are  similar  terms.  3  ab  and  14 mn  are  dissimilar 
terms. 

29.  The  common  expressions  of  algebra  are  frequently- 
named  in  accordance  with  the  number  of  terms  composing 
them.     The  following  names  are  generally  used  : 

A  Monomial.     An  algebraic  expression  of  one  term. 

4  a,  5  win,  —  Sxy,  and  17  xyz  are  monomials. 
A  Binomial.     An  algebraic  expression  of  two  terms. 

a  +  b,  3  m  —  x,  10  —  7  ay,  —  4  mnx  +  11  z  are  binomials. 
A  Trinomial.     An  algebraic  expression  of  three  terms. 

Sx  —  7  m  +  8,  4  ab  —  11  ac  —  10  mny  are  trinomials. 
A  Polynomial.     Any  expression  having  two  or  more  terms. 

While  the  binomial  and  the  trinomial  both  come  under  this 
head  they  occur  so  frequently  that  common  practice  gives  each 
a  distinct  name.  Expressions  having  four  or  more  terms  are 
ordinarily  named  polynomials. 

Oral  Drill 

Kead  the  following  algebraic  expressions : 

1.  a  +  3x  —  4:mn-\-cdn  —  3xy. 

2.  2  mx  —  3  acd  -f-  (a  -f-  x)  —  (m  -f-  w). 

3.  (a  —  x)  —  (c  —  d)  +  (m  —  y  +  z). 

4.  5ayz  —  2(2  m  —  n)  -f-  a(a  —  x)  -f  3  a(2 a  —  3y). 

5.  —  mnx  +  3  a(c  —  2  d  + 1)  —  (a  —  m  +  n)sc  -f- 12(4  —  sc). 

6.  (a -X)(c  +  y)-(x  +  2)0/ - 3) - ■(«+  l)(a>  +  2)(*  +  3). 

&     2/     3(c-d)  "c-2/ 


CHAPTER   II 
ADDITION.    PARENTHESES 

30.  Addition  in  algebra,  as  in  arithmetic,  is  the  process  of 
combining  two  or  more  expressions  into  an  equivalent  ex- 
pression or  sum.  The  given  expressions  to  be  added  are  the 
addends. 

THE  NUMBER  PRINCIPLES  OF  ADDITION 

31.  The  Law  of  Order.     Algebraic  numbers  may  be  added  in 

any  order. 

In  general :  '  a  +  6  =  b  +  a. 

Numerical  Illustration :  5  +  3  =  3  +  5. 

32.  The  Law  of  Grouping.  The  sum  of  three  or  more  alge-, 
braic  numbers  is  the  same  in  whatever  manner  the  numbers  are 
grouped. 

In  general:  a+b+c=a+(b+c)  =(a+6)  +  c=(a+c)  +  6. 

Numerical  Illustration:  2+3 +4 =2  +(3 +4)  =  (2 +3)  +4=  (2 +4)  +3. 

A  rigid  proof  of  these  laws  is  not  necessary  at  this  point ; 
but  may  be  reserved  for  later  work  in  elementary  algebra. 

The  law  of  order  is  frequently  called  the  commutative  law, 
and  the  law  of  grouping  is  called  the  associative  law. 

ADDITION  OF  MONOMIALS 

The  principles  underlying  the  addition  of  the  simplest  forms 
of  algebraic  expressions  have  already  been  developed,  and  they 
are  readily  applied  in  the  more  difficult  forms  of  later  work. 

22 


ADDITION   OF  MONOMIALS  23 

(1)   The  sum  of  like  quantities  having  the  same  sign,  all  -f  or 
all  - . 


By  Articles  22  and  23  : 

+  7 

+  12a 

-7 

-12a 

+  3 

+    5a 

-3 

—   5a 

+  10 

+  17  a 

-10 

-17  a 

In  general : 

33.  The  coefficient  of  the  sum  of  similar  terms  having  like 
sigiis  is  the  sum  of  the  coefficients  of  the  given  terms  with  the 
common  sign. 

(2)  The  sum  of  like  quantities  having  different  signs,  some  + 
and  some  — . 

By  Article  24 :  +7  +  12  a  -  7  -  12  a 
-3  -  5q  +  3  +  5q 
+  4        +    la        -4        -    la 

In  general : 

34.  The  coefficient  of  the  sum  of  similar  terms  having  unlike 
signs  is  the  arithmetical  difference  between  the  sum  of  the  +-  co- 
efficients and  the  sum  of  the  —  coefficients,  with  the  sign  of  the 
greater. 

Oral  Drill 


Add 

orally : 

1. 

5a 
Sa 

2. 

6a 
9a 

3. 

4a; 

7x 

4. 
9mn 
Smn 

5. 

Sbcd 
11  bed 

6. 

7 amx 
19  amx 

7. 
17  cmy 
19  cmy 

8. 
-4a6 
-3a6 

9. 

—  8  en 
—5  en 

10. 

—  Sxz 
—xz 

11. 

—  15cz 
-11  cz 

12. 

—Sexy 
— 14:  cxy 

13. 

-12  abx 
—19  abx 

14. 

—  5  dny 
—  dny 

15. 

10  ac 
—6ac 

16. 

— 19  mx 
9mx 

17. 

-Scy 
15  cy 

18. 
12  ax 

—  Sax 

19. 

— 15  cz 
15  cz 

20. 

27  axy 

— 16  axy 

21. 

—  Icdmn 
45  cdmn 

24  ADDITION.     PARENTHESES 

If  the  sum  of  three  or  more  terms  is  required,  we  apply  the 
law  of  grouping  (Art.  32),  and  separately  add  the  +  and 
the  —  terms. 

Thus,  5-8  + 11 -16  +  3  =  5+ 11 +3-8-16 

=  19-24 
=  —  5.    Result. 

Let  the  student  apply  this  principle  in  the  following : 

Oral  Drill 
Add  orally : 

1.  2-7  +  6-9.  7.  —  8x  +  9x-6x  +  5x. 

2.  -8  +  3  —  12  +  7.  8.  -llac  +  14ac-ac-2ac. 

3.  —9  +  8  —  15  —  11.  9.  3 mnx  —  8  mnx  —  9 mnx  + 13 mnx. 

4.  7-14  —  3  +  10.  10.  -1  cy-§cy  +  3ey-llcy. 

5.  13-18  +  7-21.  11.  3a-5a  +  8a-lla-3a  +  a. 

6.  3  a  —  4a  +  6 a  —  3a.  12.  —  4cxy-\-3xy—7xy+xy—xy-{-8xy. 

13.  5  am  —  8  am  —  24  am  +  13  am  —  am  +  6  am  — 11  am. 

14.  6  m  —  7  —  4  m  + 11  —  5  m  — 17  —  m  +  5  ra  + 13. 

15.  —  4  ex  +  8  ex  —  3  ex  +  2  ex  —  3  ex  +  c#  — 15  ex  — 14  'ex. 

ADDITION  OF  POLYNOMIALS 

The  principles  established   for  the  addition  of  monomials 
apply  directly  to  the  addition  of  polynomials. 
Illustrations : 

1.  Find  the  sum  of  5a  +  76  — 2c,  2a  — 36  +  8  c,  and 
-3a  +  26-10c. 


In  the  customary  form  :        5a  +  76—    2c 

2a-36+    8c 

-3a  +  26-10c 

4a  +  6&-    4c    Result. 

For  the  sum  :         the  coefficient  ofa=      5  +  2—    3  = 
the  coefficient  of&=      7-3+2  = 
the  coefficient  ofc=-2+8-10  = 

4, 

6, 

-4. 

ADDITION  OF  POLYNOMIALS  25 

It  frequently  happens  that  not  all  of  the  terms  considered  are  found  in 
each  of  the  given  expressions,  in  which  case  we  arrange  the  work  so  that 
space  will  be  given  to  such  terms  as  an  examination  shows  need  for. 

2.  Add  4a  +  36  +  3m,  2b +  3 c  —  d,  2a  +  3d  +  2m  —  x, 
and  5b  — 5m  — 3 x. 

4a-f   36  +  3m 

+    26  +  3c-     d 
2a  +3d+2m-    x 

+    5  6 —  5  m  —  3% 

6a  +  10  6  +  3c  +  2(Z  -4  a;    Result. 

The  coefficient  of  the  wi-term  in  the  sum  being  0,  that  term  disappears. 

In  general,  to  add  polynomials : 

35.  Write  the  given  expressions  so  that  similar  terms  shall  stand 
in  the  same  columns.  Add  separately,  in  each  column,  the  positive 
coefficients  and  the  negative  coefficients,  and  to  the  arithmetical 
difference  of  their  sums  prefix  the  proper  sign. 

Collecting  terms  is  another  expression  for  adding  two  or  more 
given  expressions. 

36.  Checking  results.  The  accuracy  of  a  result  may  be 
checked  by  substituting  convenient  numerical  values  for  each 
of  the  given  letters.  The  substitution  is  made  both  in  the 
given  expressions  and  in  the  sum  obtained,  and  the  work  may 
be  considered  accurate  if  both  results  agree. 

Illustration  :  To  check  the  sum  of  5a  +  76  +  2c  and  2a  —  36  —  5c, 
let  a  =  1,  6  =  2,  and  c  =  1. 

Then  5a  +  76  +  2c  =  5  +  14  +  2=21 

and  2a-36-5c  =  2-    6-5  = -9 


Whence,                      7  a  +  4  6  - 

-3c  =  7+    8- 

-3 

=  12. 

• 
Exercise  2 

Find  the  sum  of : 

1.                           2. 

3. 

4. 

3a  +  76            _7a_f9 

4  n  —  9  cz 

—  acm-f-   3xyz 

5a-36               12^-1 

—  n  -f-  9  cz 

2  acm—YI  xyz 

26  ADDITION.    PARENTHESES 


5. 

6. 

7. 

5ab- 

-  7  ac  —    ac? 

-   2cx-16 

m  —    nx              —z 

11  ab 

+  2  ad 

ay  +  20  c# 

3nx  —  5my  +  z 

8.  4a+3  6,  5a  —  2  6,  — 7a+4  6,  and  4  a  —11  6. 

9.  3a  —  2c  —  jb,  4a  —  c  +  7 x,  and  —  a  —  5c+9sc. 

10.  4  x  +  3  2/  — 11,  —  5  a  —  2  y  —  8,  and  x  + 19. 

11.  4  m  —  2  n  —  sc,  —  3  n  —  4  a?,  2  m  +  3  n,  and  7  %  —  9  x. 

12.  3a  — 6  — 4c  +  7,  a  — 4  6  — c+6,  and —2a— 6  + 6c— 14. 

13.  — 3a+76— 2c  — 7,  c— 3a— 6+8,  and  7—4  6  — a +  3  c. 

14.  a  —  3y  +  m-2x  +  7,   36-2?/  +  2a-4,   7x  —  2a  —  y, 
and66  —  5a-2a-12. 

Collect  similar  terms  in : 

15.  5a6  +  66c  —  3  am  —  8a;  +  4a6  —  ac-f  3cd  —  am 

+  3ab—2cd—x. 

16.  4m  —  9  +  d  —  2/ +  3  —  2d  —  6  —  5m  —  3y-\-2m 

-y-d+16. 

17.  4a6c—  8  6cd  —  5cd*a;  — 18  +  13  6cd*  —  18a6c—  5cdx 

-7  +  12a6c. 

18.  —  9  dx  —  mn  —  3  ay  +  4  6w  — 19  —  14  ay  +  5  bn  —  15  mn 

+  8  da;  — 10  bn  + 14  mn  + 16  ay. 

PARENTHESES 

37.  The  parenthesis  is  used  in  algebra  to  indicate  that  two 
or  more  quantities  are  to  be  treated  as  a  single  quantity. 

Thus,  a  +  (b  +  c)  is  the  indicated  sum  of  a  and  the  quantity,  b  +  c. 
a  —  (6  +  c)  is  the  indicated  difference  between  a  and  the  quantity,  b  +  c. 

38.  A  sign  +  or  —  before  a  parenthesis  indicates  an  opera- 
tion to  be  performed. 

a  +  (6  4-  c)  is  an  indicated  addition, 
a  —  (6  +  c)  is  an  indicated  subtraction. 


PARENTHESES  27 

The  Parenthesis  preceded  by  the  Plus  Sign 

Consider  the  expression  20  +  (10  +  5). 

By  first  adding  10  and  5,  20  +  (10  +  5)  =  20  +  15  =  35. 

Adding  separately,  20  +  (10  +  5)  =  20  +  10  +  5=  35. 

The  result  is  clearly  the  same  whether  the  parenthesis  is  removed 
before  or  after  adding. 

Again,  consider  the  expression       20  +  (10  —  5). 

By  first  subtracting  5  from  10,       20  +  (10  -  5)  =  20  +  5  =  25. 

Or,  subtracting  separately,  20  -f  (10  -  5)  =  20  +  10  -  5  =  25. 

And,  again,  the  same  result  from  each  process. 

In  general  symbols : 

For  the  +  parenthesis  :  a  +  (b  +  c)  =  a  +  b  +  c. 
a  +  (&  —  c)  =  a  +  b  —  c. 

In  general : 

39.  If  a  +  (  )  is  removed,  the  signs  of  its  terms  are  not  changed. 

The  Parenthesis  preceded  by  the  Minus  Sign 

Consider  the  expression  20  —  (10  +  5). 

By  first  adding  10  and  5,  20  -  (10  +  5)  =  20  -  15  =  5. 

Or,  subtracting  separately,  20  —  (10  +  5)  =  20  —  10  —  5  =  5. 

The  result  is  the  same  from  each  process. 

Again,  consider  the  expression     20  —  (10  —  5). 

In  this  expression  we  are  not  to  take  all  of  10  from  20,  only  the  differ- 
ence between  10  and  5,  i.e.  (10  —  5),  being  actually  subtracted.  There- 
fore if  all  of  10  is  subtracted,  we  must  add  5  to  the  result. 

By  first  subtracting  5  from  10,      20  -  (10  -  5)  =  20  -  5  =  15. 

Or,  separately,  20  -  (10  -  5)  =  20  -  10  +  5  =  15. 

And,  again,  the  same  result  from  each  process. 

In  general  symbols : 

For  the  —  parenthesis :  a  —  (b  +  c)  =  a  —  b  —  c. 
a  —  (b  —  c)  =  a  —  b  +  c. 
In  general : 

40.  Ifa—()  is  removed,  the  sign  of  every  term  in  it  must  be 
changed. 


28 


ADDITION.     PARENTHESES 


Two  important  principles  must  be  kept  constantly  in  mind : 

(1)  The  sign  before  a  parenthesis  indicates  an  operation  of 
either  addition  or  subtraction. 

(2)  The  sign  of  a  parenthesis  disappears  when  the  parenthesis 
is  removed. 

For  example : 

15  -  (7  -  4  +  2)  =  15  -  7  +  4  -  2. 

The  —  sign  of  7  in  the  result  is  not  the  original  —  sign  of  the  paren- 
thesis. The  original  sign  of  7  was  + ,  and  this  sign  was  changed  to  —  by 
the  law  of  Art.  40. 

The  sign  of  a  parenthesis  shows  the  operation,  and  disappears 
when  we  perform  it. 

Oral  Drill 

Give  orally  the  results  on  the  following  : 

1.  (+5)  +  (+2).  io.    (_7a)-(+5a). 

2.  (+5)-(+2).  11.    _(_5a)-(-7a). 

3.  (+5)  +  (-2).  12.    -(+6m)  +  (-6m). 

4.  (+5) -(-2).  13.    (-9  a?) -(9  a;). 
5-    (-7)  +  (+6).  14.    (-9x)-(-9x). 

6.  (+5)-(-9).  15.    -(-19ran)-(-20ran). 

7.  (-10) -(-19).  16.    -(Jxyz)-(-Sxyz). 

8.  (+16)  + (-17).  17.  '(-16  6c)-(-116c). 

9.  (-14) -(-13).  18.    -(±2xy)-(-15xy). 

19.  (_5)-(-2)-(-3)  +  (-4). 

20.  a-(4a)  +  (-4a)  +  (-a). 

21.  -5x  +  (-2x)-(-x)-(3x)  +  (-x). 

22.  _(-l)-(l)_(_l)_(l)  +  (-l)-(-l). 

23.  2-(+3)-(-4)  +  (-6)-(-6). 

24.  -(-2a:)-(+3a>)-(2as)-(-3a!)  +  (-2a>). 

25.  —  (—  mn)  +  (—  3  raw)  —  (—  4  ran)  +  (—  raw). 

26.  2  a?-(- 3  «)-(+2o;)  +  (- 3  *)-(  +  »)  +  (-«). 


PAKENTHESES  29 

Parentheses  within  Parentheses 

It  is  frequently  necessary  to  inclose  in  parentheses  parts  of 
expressions  already  inclosed  in  other  parentheses,  and  to  avoid 
confusion  different  forms  of  the  parenthesis  are  used.  These 
forms  are:  (a)  the  bracket  [  ],  (6)  the  brace  {  \,  (c)  the  vin- 
culum —  .  Each  has  the  same  significance  as  the  paren- 
thesis. 

That  is,      (a  +  &)  =  [«  +  6]  =  {a  +  b]  =  a  +  b. 

In  removing  more  than  one  of  these  signs  of  aggregation 
from  an  expression : 

41.  Remove  the  parentheses  one  at  a  time,  beginning  either 
with  the  innermost  or  with  the  outermost.  If  a  parenthesis  pre- 
ceded by  a  minus  sign  is  removed,  the  signs  of  its  terms  are 
changed.     Collect  the  terms  of  the  result. 

Illustration : 


Simplify  4m- [3 m-(n+2)-4]-Sm+3n—(7i-l)j+n. 


4  m  -  [3  m  -  (n  +  2)  -  4]  -  {m  +  3  n  -  (n  -  1)}  +  n  = 
4m- [3 m  -  n-2  -  4]  -  {m  +  3 n  —  n  +  1}  +  n  = 
4m-  3m  +  n  +  2  +4  -{m  +  Sn  -  ra  +  1}  +w  = 
4m—  3  m  +  n  +  2  +4  —  m  —  3  n  +  n  —  1     +  n  =  6.     Result. 

Exercise  3 

Simplify : 

1.  4a+(3a-8).  7.  9 a +  [3 a; -(a -2)]. 

2.  5x-(3x  +  7).  8.  9z-[3a+-(a;+-2)]. 

3.  2c+(c-8)-6.  9.  ±a-\a-(-a  +  7)\. 

4.  3m  — (2  m  — ?i+-3).  10.  2ac  —  (ac  —  7  +  3ac). 


5.  4z-(8-3z)-ll.        11.    5n+[9-{3n-n  +  2\]. 

6.  —  5m—  (— m—  p-6).    12.    —  6  —  (— xy+[  — 2  xy—  (xy+3)~\). 

13.  -3a-[36-f2a-(6&-a)n. 

14.  m  —  [m  +  \m—  (m  —  m  —  1)}]. 


30  ADDITION.     PARENTHESES 


15.  -(2m  +  l)-(-[-m  +  m-3]). 

16.  7x-[2+(-3x-)x-(-6x-x-7)-x\)]. 


17.    _(_a_i)_2a-(a  +  l)-3a-5-8. 


18.    l_(_l)+)_l-lH-a-(-l)i, 


19.    2-(-l)+j-l-l  +  a-(-l)-(-a)J. 


20.    -H-(_l)_{l  +  l-(a  +  l)+(-l)t. 


21.   3m- [2m—  \m  — n  —  3m— 9— 2  m— 7+3  nj  —  (— m— 5)]. 


22.   2a-(36  +  c)-2a  +  (c-36) 

-Jc-(2a-36)-[a-(6-c)] 


23.  l_(l+a)-[-j-[(-l-a-l-3a)-a]-lS-a]. 

24.  ax— (-  aa;)—  |  —  (—ax)\  —  [  —  (—\  —  (  —  ax)\)]. 

25.  (m-  1)—  j— (—  J  —  (—  m  +  1)  —  nj  —  m)—  raj. 

Inclosing  Terms  in  Parentheses 

This  operation  is  the  opposite  of  the  removal  of  parentheses, 
hence  we  may  invert  the  principles  of  Arts.  39  and  40  and 
obtain : 

42.  Any  number  of  terms  may  be  inclosed  in  a  +  (  )  without 
changing  the  signs  of  the  terms  inclosed. 

43.  Any  number  of  terms  may  be  inclosed  in  a  —  (  )  provided 
the  sign  of  each  term  inclosed  be  changed. 

Illustrations : 

(Art.  42. )     1.  a  +  b  +  c-d  +  e  =  a  +  b  +  (c-d+e). 

(Art.  42.)     2.  a  +  6-c-d  +  e  =  a  +  &  +  (-c-d  +  e). 

(Art.  43.)     3.  a  +  6-c  +  d-e  =  a  +  &-(c-d  +  e). 

(Art.  43.)    4.  a-b  +  c-d  +  e  =  a-b-(-c  +  d-e). 


PARENTHESES  31 

Exercise  4 

Inclose  the  last  two  terms  of  each  of  the  following  in  a 
parenthesis  preceded  by  the  plus  sign  : 

1.  a  +  3  6-|-9  c.  4    12  mn  +  4  mx  —  5  my  -f- 16. 

2.  4mH-3n-f-2j9  — 8.  5.   5c  —  5d  —  x  —  mny. 

3.  4ae  +  6ad  +  9ae-ll.       6.    -9c  +  lld~6  +  a». 

Inclose  the  last  three  terms  in  a  parenthesis  preceded  by 
the  minus  sign : 

7.  4a-56  +  3c-9.  11.  3x  +  ±y  +  5z  +  2. 

8.  2a  —  66  —  c  +  x.  12.  4 ac  —  5 6c  +  3  bd  + 10  ad. 

9.  5a;  — 8y  +  3«  — 11.  13.  12  —  w  +  ra  —  2p  +  x. 
10.    —  7  —  5ra  +  3w  —  2  p.  14.  m  +  n—p  +  x  —  y  +  z. 

15.    —  2  —  x  -f  a»  —  xy  —  yz. 

Without  changing  the  order  of  the  terms,  write  each  of  the 
following  expressions  first  in  binomials  and  then  in  trinomials. 
Before  each  parenthesis  use  as  its  sign  the  sign  of  the  term 
which  is  to  be'  first  in  that  parenthesis : 

16.  a-f-6  —  c-fd  —  x  —  y. 

17.  a  —  c  +  d  —  m  —  n  +  z. 

18.  c  —  ra  —  x  +  y  —  z  —  1. 

19.  a  —  d  —  e  —  m  —  n  +  x. 

20.  m  —  n—p  —  x  —  y  —  z. 

21.  ab  —  ac  +  cd  —  mn  —  np  +  mp. 

22.  2a  +  3c-f  4a,  +  5a  —  3y-2z. 

23.  7a6  —  3ac  +  4  6d'  —  5 6c +  3 ad  —  4ca\ 

24.  5ay  —  3az  +  8z  —  4:xy  —  3xz  +  2yz. 

25.  ama;  —  cny  —  bz  —  cnx  —  amy  —  dz. 


CHAPTER   III 
SUBTRACTION.    REVIEW 

44.  Subtraction  is  the  inverse  of  addition.  In  subtraction 
we  are  given  the  algebraic  sum  and  one  of  two  numbers,  and 
the  other  of  the  two  numbers  is  required.  The  given  sum  is 
the  minuend,  the  given  number  the  subtrahend,  and  the  required 
number  the  difference  or  remainder. 

45.  We  may  base  the  process  of  subtraction  on  the  principle 
of  Art.  40,  for, 

The  quantity  to  be  subtracted  may  be  considered  as  inclosed  in 
a  parenthesis  preceded  by  a  minus  sign.     Consider  the  example, 

From  10a  +  36  +  7c  take  6  a  +  b  -  c. 

By  definition,  the  first  expression  is  the  niinuend,  the  second 
the  subtrahend.  Inclosing  both  expressions  in  parentheses, 
and  replacing  the  word  "  take "  by  the  sign  of  operation  for 
subtraction,  we  have 

(10  a  +  3  b  +  7  c)  -  (6  a  +  6  -  c). 

Removing  parentheses,    10a  +  36  +  7c  —  6a  —  b  +  c. 

Collecting  terms,  4  a  +  2  b  +  8  c.     Result. 

We  have,  therefore,  changed  the  signs  of  the  terms  of  the  sub- 
trahend and  added  the  resulting  expression  to  the  minuend.  In 
practice  the  usual  form  would  be 

Minuend  :  10  a  +  3  b  +  7  c. 

Subtrahend :  6  a  +    b  —     c. 

Difference :  4a  +  2&  +  8c.    Result. 

The  change  of  signs  should  be  made  mentally.  Under  no  cir- 
cumstances should  the  given  signs  of  the  subtrahend  be  actually 
altered. 

32 


SUBTRACTION  33 

From  the  foregoing  we  make  the  general  statement  for  sub- 
traction of  one  algebraic  expression  from  another. 

46.  Place  similar  terms  in  vertical  columns.  Consider  the  sign 
of  each  term  of  the  subtrahend  to  be  changed,  and  proceed  as  in 
addition. 

Two  important  principles  result  from  the  process  of  sub- 
traction. 

47.  Subtracting  a  positive  quantity  is  the  same  in  effect  as 
adding  a  negative  quantity.     In  general : 

By  Art.  40:  (+  a)  -(+&)  =  «-&.     By  Art.  39  :  («j-a)  +  (- 6)=a-&. 
Numerical  Illustrations : 
(+  5)  -  (+  3)  =  5  -  3  =  2.  (+  5)  +  (-  3)  =  5  -  3  =  2. 

48.  Subtracting  a  negative  quantity  is  the  same  in  effect  as 
adding  a  positive  quantity.     In  general : 

By  Art.  40:  (+«)-(-  &)  =  «  +  b.    By  Art.39:  (+  a)  +  (+  b)=a  +  b. 

Numerical  Illustrations : 
(+  6)  -  (-  3)  =  5  +  3  =  8.  (+  5)  +  (+  3)  =  5  +  3  =  8. 

Oral  DriU 


From 

1. 

13 

2. 

12  a 

3. 

-9a 

4. 
-3a 

5. 

13  ab 

6. 
15  a 

Take 

_4 

5  a 

-2a 

—  6  a 

-1  ab 

-3x 

From 

7. 
9m 

8. 

-7c 

9. 

1  xy 

10. 

—  Sac 

11. 

— 19  mn 

12. 

—  7  xz 

Take 

13  m 

8c     - 

-5xy 

—  Sac 

—  11  mn 

23  xz 

From 

13. 

—  lac 

14. 

+  7ac 

15. 

0 

16. 

0 

17. 

abc 

18. 

—  4  acx 

Take 

-f-7ac 

—  7ac 

9x 

-9a 

—  9abc 

acx 

SOM.    EL.    ALG. 


34  SUBTRACTION.    REVIEW 

SUBTRACTION  OF  POLYNOMIALS 

49.  By  application  of  the  principle  of  Art.  46,  we  subtract 
one  polynomial  from  another.  Results  in  subtraction  may  be 
tested  as  shown  in  Art.  36. 


Exercise  5 

1. 

From            5  a  +  7 
Take             3a-4 

2. 

8m-19 
3  m-  11 

3. 

-  7  c  + 14 
-9c  +  14 

4. 
-17ac-8 
-   9ac  +  9 

5. 

From  3a  +  46*-7 
Take             3  6-9 

6. 

6x-$y 
x  +  3y- z 

7. 
-3m 
2  a  — 5  m 

+  7 

8. 
0 

a  +  b  —  c 

9.  From  7  a  +  3  m  —  9  take  2  a  —  5  m  4-  8. 

10.  From  16a-|-6a;—  3y  take  a  —  7  x  — 15 y. 

11.  From  4a  +  9a"-18  take  -3a  +  9d'-15. 

12.  From  5x—7y-\-3z  take  3y  —  1  z. 

13.  From  — 7a  +  6c-r-3cZ  +  5  take  2  a  —  5  c  +  3. 

14.  From  11  a  +  3  m  —  a;  take  2  a  +  7  #  —  m. 

15.  From  26  +  5c  —  3n  take  —  2  a  —  6  +  4  c  —  3  n. 

16.  Subtract  15  aa?  —  3  ay  —  19  from  17  ax  —  hay  —  11. 

17.  Subtract  10  —  3x  —  1  y  —  z  from  z  +  ll  —  2x-3y. 

18.  Subtract  5c  +  6d  —  4m  +  n  from  3c  —  m. 

19.  Subtract  5a  — 11  6  —  2c  from  3a  —  8  c  +  2  ra. 

20.  From  the  sum  of  3m  —  n  +  2p  and  3m-4n-5|)  take 
the  sum  of  m  -f-  3  n  —p  and  3  m  —  7  n  +  6 p. 

21.  From  the  sum  of  5x—  3y  +  2  z  -f  3  and  3  x  +  7  subtract 
the  sum  of  3 a;  —  y 4- 2  and  7  —  2x  —  y-\-llz. 

22.  Subtract  the  sum  of  4  a  —  11  c  +  d  and  3  b  —  c  +  10  from 
the  sum  of36  +  d  —  8c  and  4  a  —  4  c  +  9. 

23.  Take  the  sum  of  3  m— n+2y,  2n— 3  m— 4^  and  m+p— z 
from  the  sum  of  m  —  3 3,  3 n  +p,  and p-f-m  —  2y  +  n  —  z. 


GENERAL  REVIEW  35 

50.  Addition  and  Subtraction  with  Dissimilar  Coefficients. 
When  the  coefficients  of  similar  terms  are  themselves  dis- 
similar, the  processes  of  addition  and  subtraction  result  in 
expressions  with  compound  coefficients,  i.e.  coefficients  having 
two  or  more  terms.  The  principle  is  easily  understood  from 
the  following  illustrations : 

Addition  Subtraction 

ax  cm  am  cdz 

bx_  —  acm  cm  —  9  amdz 

(a  +  b)x  (1  —  a)  cm  (a  —  c)m        (c  +  9  am)  dz 

Exercise  6 

Find  the  sum  of : 

1.  am  +  dm.  4.    —  17  mxy  +  2pxy.     7.   bcdx  +  12  cdx. 

2.  cdx  +  5  nx.  5.    cdn  —  7  dn.  8.   5  cdx  —  12  bdx. 

3.  3  bed  —  2  mcd.     6.   acx  — 19  bx.  9.    —  mxy  — 11  cxy. 

10.  6a;  +  ay  and  ma;  +  ny.  13.   5  ab  —  2  cd  and  3  6  +  7  a\ 

11.  a#  +  nz  and  6aj+pz.  14.   6  az  -f-  5  by  and  raz  —  3  y. 

12.  acm  +  6n  and  dm  +  cfri.      15.   11  mxy  +  5  ccz  and  ascy  —  acz. 

16.  2 ftm  + 3 a; +  ?/ and  5m +cx  —  m?/. 

17.  3  a6  +  7  ac  + 11,  m&  —  lie  —  7,  and  5  6  —  nc  +  3. 

18.  (3  +  c)x +  (2  c  -  9)a>  +  (6  -  3  c)x  +  (a  +  c  +  1)*. 

(Examples  1-12  inclusive  will  serve  as  an  exercise  for  subtraction  ;  the 
first  expression  in  each  being  the  given  minuend,  the  second  expression, 
the  subtrahend.) 

GENERAL  REVIEW 

Exercise  7 

1.   Find  the  sum  of  3  a  +  2  6  —  6  m,  2  b  —  5  c  +  x,  —  4  a—  76 
-f  c  —  m,  and  a  —  3  c  +  x. 


2.  Simplify  1-J1- [1- (1-1  + m)]-mj  -m. 

3.  Add   3a  +  26-c,   2a-36  +  3c,   4a  +  36-7c,   and 
-8a-46  +  5c. 


36  SUBTRACTION.     REVIEW 

4.  Subtract  4  x  -f-  3  y  —  2  from  5  as,  and  add  x  —  3  y  +  2  to 
the  result. 

5.  Subtract  2#  —  3y  from  4aj+"2y,  subtract  this  result 
from  7  x  —  Sy,  and  add  a;  —  ?/  to  the  final  remainder. 

6.  What  expression  must  be  added  to  4a  —  Sm-\-x  to  pro- 
duce la  —  5  ra  +  3  a;  ? 

7.  What    is   the   value   of    (-3)  -  (+  2)  -  (-3)  +  (-  5) 
-(-2)? 

8.  Collect  —  3a  —  &  +  4  c,  2c  —  3  a*  -f  #,  a  —  4a—  d,  5  c +  2  a 
+  3  as,  and  4  d  +  6  -  7  c. 


9.    Simplify  x  —  1  —  J#  —  1  —  [a  —  1  —  a;  —  1  —  ay } . 

10.  Subtract  4  a?  —  3  y  + 11  from  unity,  and  add  5  a;  —  3  y-h  12 
to  the  result. 

11.  Combine  the  m-terms,  the  n-terms,  and  the  avterms  in  the 
following,  inclosing  the  resulting  coefficients  in  parentheses : 
am  -{■  3  bn  -\-  x  +  m  +  n  —  ax  +  2  n  —  ex. 

12.  A  given  minuend  is  7  x  + 12  y  —  7,  and  the  corresponding 
difference,  4  x  —  y  +  2.     Find  the  subtrahend. 

13.  To  what  expression  must  you  add  2a  — 3c  +  m  to  pro- 
duce 5  a  -f  7  c  —  9m? 

14.  Subtract  2a—7x+3  from  the  sum  of  3  x  +  2  a,  4  a 
— 10  #,  5  a  —  7,  and  —  4  a;  — 11  a. 

15.  From  (a  +  c)  y  +  (m  -f-  n) z  take  (a  —  c)i/  —  (m  —  n)z. 

16.  Subtract  a;  +  17  y  —  2  from  12  x  -f-  3  z  and  add  the  result 
to3x+(y  +  llz)-4y. 

17.  Add  the  sum  of  4  a;  +  7  c  and  2  a?  +  3  c  to  the  remainder 
that  results  when  x  -f  4  c  is  subtracted  from  5  c  —  11  x. 

18.  From  (a  +  4)  x  +  (a  -f  3)  y  subtract  (a  + 1)  x  -f  (a  +  2)  ?/. 

19.  Add  3 a?  +  2 m  —  1  and  2x  —  3m-\-7,  and  subtract  the 
sum  from  6  a;  —  m  +  6. 


GENERAL   REVIEW  37 


20.  Simplify  a  —  1  —  a  + 1  +  [a  —  1  —  a  —  a  —  1  —  (a  —  1)  — 

21.  From   7ra  +  3<c  — 12    take    the    sum    of    3  a; +  7    and 
2m-3y-3. 

22.  What  expression  must   be   added  to  a  +  b  +  c  to  pro- 
duce 0  ? 

23.  What  expression   must   be  subtracted   from  0  to  give 

a  +  6  +  c? 

24.  From   what    expression    must    4  x  —  7  m  + 10  be   sub- 
tracted to  give  a  remainder  of  3  #  -f  6  ra  —  4  n  -f  2  ? 

25.  Inclose  the  last  four  terms  of  a  —  86-f-4m  —  5n  +  7  x 
in  a  parenthesis  preceded  by  a  minus  sign. 

26.  Simplify  and  collect  a  +  [  —  2m  —  4  a  +  <c  — 

(—  2  <c  —  m  -f-  a)  —  3  x\ 

27.  If  x  +  7y  —  9  is   subtracted   from   0,  what  expression 
results  ? 

28.  Isthesumof  [-7  +  (-2)-(-3)]  +  [-!-2  +  (-l)i] 
positive  or  negative  ? 

29.  Prove  that  [3- (-2) -(-!)]  +  [(- 2) +  (- 1)_(_3)] 

+[_2_(_l)  +  (_5)]=0. 

30.  Simplify  and  collect 

10_[9-8-(7-6-{5-4-3^2S-l)]. 

31.  Show  that 

2  -  ( -  3  +  o^l)  -  3  +  ( -  2  -  (Tfl )  =  _ 2  a. 

32.  Collect  the  coefficients  of  x,  y,  and  z  in 

abx  —  acy  +  abz  —  mnx  +  mpy  —  npz  +  x  —  y  —  z. 

33.  Simplify  and  collect 

l-[-S-(-l  +  r^)-lf-l]-a. 

34.  What  expression  must  be  subtracted  from  a  —  x  +  3  to 
give     -(a  +  [-«-(-2a  -x  + 1)-  3])  ? 


CHAPTER   IV 

MULTIPLICATION 

51.  Multiplication  is  an  abbreviated  form  of  addition. 

Thus,  3x4  =  4  +  4  +  4.     5a  =  a  +  a  +  a  +  a  +  a. 

For  the  purpose  of  arithmetic  multiplication  has  been 
denned  as  the  process  of  taking  one  quantity  (the  multiplicand) 
as  many  times  as  there  are  units  in  another  quantity  (the 
multiplier).  This  definition  will  not  hold  true  when  the  mul- 
tiplier is  negative  or  fractional.  Hence,  the  need  for  the 
following  definition : 

52.  Multiplication  is  the  process  of  performing  on  one 
factor  (the  multiplicand)  the  same  operation  that  was  per- 
formed upon  unity  to  produce  the  other  factor  (the  multiplier). 
The  result  of  a  multiplication  is  a  product. 

Illustration : 

The  Integral  Multiplier. 

3x5  means  that  5  is  taken  three  times  in  a  sum.  Or,  3x5  =  5  +  5+5. 
By  the  same  process  the  multiplier  was  obtained  from  unity,  for 
3  =  1  +  1  +  1. 

The  Fractional  Multiplier. 

Obtained  from  unity  a  multiplier,  2f  =  l  +  l  +  £  +  £  +  £.  For  unity 
was  taken  twice  as  an  addend  and  \  of  unity  taken  three  times  as  an 
addend. 

In  like  manner,  2fx5  =  5  +  5  +  f  +  f  +  |  =  10  +  -^  =  13|. 

THE  NUMBER  PRINCIPLES  OF  MULTIPLICATION 

53.  The  Law  of  Order.  Algebraic  numbers  may  be  multiplied 
in  any  order. 

In  general :  ab  =  ba. 

Numerical  Illustration :  3x5  =  5x3. 
38 


SIGNS  IN  MULTIPLICATION  39 

54.  The  Law  of  Grouping.  The  product  of  three  or  more 
algebraic  numbers  is  the  same  in  whatever  manner  the  numbers 
are  grouped. 

In  general :  abc  =  a(bc)  =  {ab)c  =  (ac)b. 

Numerical  Illustration :  2  . 3  .  5  =  2(3  .  5)  =  (2 :  3)5  =  (2  .  5)3. 

55.  The  Law  of  Distribution.  The  product  of  a  polynomial 
by  a  monomial  equals  the  sum  of  the  products  obtained  by  multi- 
plying each  term  of  the  polynomial  by  the  monomial. 

In  general :  a(x  +  y  +  z)=  ax  +  ay  +  as. 

Numerical  Illustration :  2(3  +  4  +  5)  =  6  +  8  +  10. 
As  in  the  case  of  addition,  no  rigid  proof  of  these  laws  is 
ordinarily   required   until    the    later   practice   of  elementary- 
algebra. 

Here,  as  in#  addition,  the  law  of  order  is  frequently  called 
the  commutative  law,  and  the  law  of  grouping  is  called  the 
associative  law. 

SIGNS  IN  MULTIPLICATION 

Upon  the  definition  of  multiplication  we  may  establish  the 
results  for  all  possible  cases  in  which  the  multiplicand  or 
multiplier,  or  both,  are  negative  numbers. 

(1)  A  positive  multiplier  indicates  a  product  to  be  added. 

(2)  A  negative  multiplier  indicates  a  product  to  be  sub- 
tracted. 

(1)  The  Positive  Multiplier.  Expressed  with  all  signs  (mul- 
tiplier =  +  3) : 

(+  3)  x  (+  5)  =  +  5  +  5  +  5  =  +  15. 
(+  3)  x  (-  5)  =  -  5  -  5  -  5  =  -  15. 

(2)  The  Negative  Multiplier.  Expressed  with  all  signs  (mul- 
tiplier =  —  3)  : 

(-  3)  x  (+  5)  =  -  (+  5  +  6  +  5)  =  -  (+  15)  =  -  15. 
(_  3)  x  (-  5)  =  -  (-  5  -  5  -  5)  =  -  (-  15)  =  +  15. 


+ 

5 

-    5 

— 

3 

-    3 

— 

16 

+  16 

(- 

a)  (  +  6)  = 

—  ab. 

(- 

<*)(- 

-b)  = 

+  ab. 

40  MULTIPLICATION 

Comparing  the  four  cases  in  the  ordinary  form  of  multiplica- 
tion, we  have 

+    6  -    6 

±    3  ±   3 

+  15  -15 

In  general : 

(+«)(+ &)=  +  «&. 
(+a)(-b)=  -  ab. 

56.  Like  signs  in  multiplication  give  a  positive  result. 

57.  Unlike  signs  in  multiplication  give  a  negative  result. 

Oral  Drill 

Give  the  products  in  the  following,  each  with  its  proper 
sign : 

1.  (3)  (-6).  7.  (-6)  (-7).  13.  (2)(-5)(0). 

2.  (4)(-5).  8.  (-9)(0).  14.  (_5)(3)(-8). 

3.  (-3)  (6).  9.  (-3)  (-11).  15.  -(2)  (3)  (-4). 

4.  (-4)  (5).  10.  (-4)  (-3)  (2).  16.  -  (3)  (-5)  (-2). 

5.  (-5)  (-4).  11.  (2)  (-5)  (3).  17.  -  (-  4)(- l)(-2).' 

6.  (5)(0).  12.  (_2)(5)(-3).  18.  -(-3) (-5) (-8). 

COEFFICIENTS  IN  MULTIPLICATION 

58.  The  coefficient  of  a  term  in  a  product  of  two  algebraic 
expressions  is  the  product  of  the  coefficients  in  the  given  multiplier 
and  multiplicand. 

The  principle  is  established  by  means  of  the  Law  of  Grouping. 
For  the  coefficient  of  mn  in  the  product  of  am  times  en: 
By  Art.  64 :  am  x  en  =  (a  x  c)(m  x  n) 

=  (ac)  (mn) 

=  acmn. 
And  the  required  coefficient  is  ac. 


EXPONENTS   IN   MULTIPLICATION  41 

EXPONENTS  IN  MULTIPLICATION 

59.  An  exponent  is  a  symbol,  numerical  or  literal,  written 
above  and  to  the  right  of  a  given  quantity,  to  indicate  how 
many  times  that  quantity  occurs  as  a  factor. 

Thus,  if  three  a's  occur  as  factors  of  a  number,  we  write  a3,  and 
avoid  the  otherwise  cumbersome  form  of  a  x  a  x  a.  In  like  manner, 
axaxaxbxb=  asb2,  and  is  read  "  a  cube,  b  square." 

60.  The  product  of  two  or  more  equal  factors  is  a  power. 
Any  one  of  the  equal  factors  of  a  power  is  a  root. 

In  common  practice,  literal  and  other  factors  having  ex- 
ponents greater  than  3  are  read  as  powers. 

Thus,  a6  is  read  "  a  sixth  power,"  or  merely  "  a  sixth." 

a3y7z2  is  read  "  a  cube,  y  seventh,  z  square." 

The  exponent  "1"  is  neither  written  nor  read.  That  is,  a 
is  the  same  as  a1.  The  difference  between  coefficients  and 
exponents  must  be  clearly  understood.  A  numerical  illustra- 
tion emphasizes  that  difference.     Thus : 

If  5  is  a  coefficient,  5x2  =  24-2  +  2  +  2  +  2=  10. 
If  5  is  an  exponent,        25  =2x2x2x2x2  =  32. 

The  General  Law  for  Exponents  in  Multiplication 

By  definition,  as  =  a  x  a  x  a, 

a*  =  axaxaxa. 
Therefore,  azxa*  =  axaxaxaxaxaxa 

=  a\ 
Similarly,  ab  x  at  =  a5+4  =  a9, 

m?x  m8  xm  =  m2+8+1  =  m6. 

In  general,  therefore,  we  have  the  following : 

If  m  and  n  are  any  positive  integers : 

am  =  ax  ax  a -"torn  factors, 
an  =  ax  ax  a-"to  n  factors. 
Hence,  am  x  an  =  (a  x  a  x  a  •••  to  m  factors)  (a  x  a  x  a  •••  to  n  factors) 
=  (a  x  a  x  a  •••  to  m  +  n  factors) 
—  am+n. 
In  the  same  manner,  am  x  an  x  aP  =  am+n+P,  and  so  on,  indefinitely. 


42  MULTIPLICATION 

This  principle  establishes  the  first  index  law,  m  and  n  be- 
ing positive  and  integral. 

The  general  statement  of  this  important  law  follows : 

61.  The  product  of  two  or  more  powers  of  a  given  factor  is  a 
power  whose  exponent  is  the  sum  of  the  given  exponents  of  that 
factor. 

Oral  Drill 

Give  orally  the  products  of  the  following : 

1.  a2  X  a3.  5.    x*  xx9.  9.    c7  x  c3  X  c5. 

2.  m5  x  m3.  6.   x6  x  x23.  10.   y9  xy5  xy3. 

3.  d7  x  d3.  7.   m  X  m2  x  m3.  11.    a2  x  a3  x  a5  X  a9. 

4.  z9  X  z8.  8.    a3  X  a2  X  a.  12.    »2Xw4X  n6  X  n8. 

MULTIPLICATION  OF  A  MONOMIAL  BY  A  MONOMIAL 

By  application  of  the  law  of  order  and  the  principles  for 
signs  and  exponents,  we  obtain  a  process  for  the  multiplication 
of  a  monomial  by  a  monomial. 

Illustrations : 

1.  Multiply  3  a2b3  by  12  aAb2. 

By  the  Law  of  Order,  3  a253  x  12  a462  =  3  x  12  x  a2  x  a4  x  68  x  b2. 
By  the  Law  of  Grouping,  =(3  x  12) (a2  x  a*)(63  x  62). 

By  Arts.  58  and  61,  =  36  a6&5.     Result. 

2.  Multiply  -7  a2b5x3z  by  5  a7b2y. 

-7  a2b5x8z  x  5  aWy  =  -7  x  5  x  a2  x  a7  x  b5  x  b2  x  xs  x  z  x  y  (53) 

=  (  -  7  x  5)  (a2  x  a?)  (65  x  b2)  (x3)  («)  (y)  (54) 

=  -35a9&7z3«/z.     Result.  (57)  (61) 

Therefore,  to  multiply  a  monomial  by  a  monomial : 

62.  Observing  the  law  of  signs,  obtain  the  product  of  the 
numerical  coefficients.  The  exponent  of  each  literal  factor  in  the 
product  is  the  sum  of  the  exponents  of  that  factor  in  the  multipli- 
cand and  multiplier. 


MULTIPLICATION  OF  POLYNOMIAL  BY  MONOMIAL     43 
t 

Oral  Drill 

Give  orally  the  products  of  the  following : 


1.             2.           3. 

4.              5.                  6. 

7. 

5a         Sx     —  4=a 

5m      —   5x        —3x 

-  %y 

—  3      — 7x     —5a 

■8m      — 11  x          16  x 

-13  2/ 

8.              9.             10. 

11.                     12. 

13. 

-Sab2       4  m3n         5  x*y 

—  3  xy3         —  6  m3ny 

—  3  mna; 

2ab     -3mn3     -1  tftf 

—  IxPy       — 11  mn3y 

10  m3x 

14.   4  abc  by  3  acd. 

20.    -  c2d3m2  by  - 

5  m3n. 

15.   4  axy  by  —  7  xyz. 

21.   c3dxy  by  — 11 

cWy3. 

16.    aW  by  a3b2d. 

22.    — 10  tfyh  by  xPy3. 

17.   4  x?yz  by  —  a^/z2. 

23.    — 11  cmn3y  by 

—  5  m2n2. 

18.   3  a7  by  —  4  amn. 

24.    13  (Wa8  by  - 

2c2d?. 

19.    — 12  a3mQ  by  —  2  m3w2z.        25.   15  an2a?z  by  —  8  ?i3ra2/. 

MULTIPLICATION  OF  A  POLYNOMIAL   BY  A  MONOMIAL 

The  process  of  multiplying  a  polynomial  by  a  monomial 
results  directly  from  the  number  principle  for  multiplication 
assumed  in  Art.  55.     That  is : 

a(x  +  y  +  z)  =  ax  +  ay-\-az. 

In  common  practice  the  multiplicand  and  multiplier  are 
written  as  in  arithmetic,  excepting  that  the  multiplier  is 
usually  written  at  the  extreme  left. 

Illustration  : 

Multiply  3  m3—  5  m2n  -\-  7  raw2  —  2  ns  by  —  2  mn. 
3  w8  -  5  m2n  +  7  mn2  -2n8  Each  term  0f  the  prod- 

~~2mn uct  is  obtained  by  the 

-6  m%  +  10  m%2  -  14  w2n8  +  4  mri4  Result,  principles  of  Art.  62,  for 
the  operation  is  made  up  of  successive  multiplications  of  a  monomial  by  a 
monomial. 


44  MULTIPLICATION 

Hence,  to  multiply  a  polynomial  by  a  monomial : 

63.  Multiply  separately  each  term  of  the  multiplicand  by  the 
multiplier,  and  connect  the  terms  of  the  resulting  polynomial  by 
the  proper  signs. 

Exercise  8 

Multiply : 

1.  2.  3.  4. 

3a+7x        2a2  — 10a        llax-15ay        a^-lOx-ll 
2  a  3  a3  —  axy  x2 

5.  6.  7. 

2  ra2  —  10  ran  + 15  n2      —x*  —  x2y  +  xy2      a4  —  3  a3x  +  3  ax3  —  x* 
3n  —x*  ax 

8.  10  a  (7  ab  -  8  ac  + 11  be).        11.   3  a  (a3-  a2b-\-ab2). 

9.  —3x(x2-x  +  ll).  12.    -±bc(bx-bm  +  3bn). 
10.    -2ra2(ra3n-ra2n2  +  ran8).      13.    11  a2»(^  +  9»- 15). 

64.  The  degree  of  a  term  is  determined  by  the  number  of 
literal  factors  in  that  term. 

7  a8x2  is  a  term  of  the  5th  degree,  for  3  +  2  =  5. 

65.  The  degree  of  an  algebraic  expression  is  determined  by 
the  term  of  highest  degree  in  that  expression. 

5  m2n  +  mn  +  n2  is  an  expression  of  the  3d  degree. 

66.  An  algebraic  expression  is  arranged  in  order  when  its 
terms  are  written  in  accordance  with  the  powers  of  some  letter 
in  the  expression. 

If  the  powers  of  the  selected  letter  increase  from  left  to  right,  the 
expression  is  arranged  in  ascending  order. 

Thus,  x  -  2  z2  +  5  x*  -  7  a4  +  10  x5. 

If  the  powers  of  the  selected  letter  decrease  from  left  to  right,  the 
expression  is  arranged  in  descending  order. 

Thus,  4  x9  -  5  x7  +  3  x*>  -  2  zs  -  3  x. 


MULTIPLICATION    OF  POLYNOMIAL  BY   POLYNOMIAL      45 

67.  The  degree  of  a  product  is  equal  to  the  sum  of  the  degrees 
of  its  factors. 

68.  A  polynomial  is  called  homogeneous  when  its  terms  are 

all  of  the  same  degree. 

Thus,     sc*  —  4  xsy  +  6  x2y2  —  4  xyz  +  y4  is  a  homogeneous  polynomial. 

MULTIPLICATION  OF  A  POLYNOMIAL  BY  A  POLYNOMIAL 

A  further  application  of  the  law  of  distribution  for  multipli- 
cation (Art.  55)  establishes  the  principle  for  multiplying  a 
polynomial  by  a  polynomial. 

By  Art.  55 :       (a  +  b)  (as  +  y)  =  a  (x  +  y)  +  b  (x  +  y) 

=  ax  +  ay  +  bx  +  by. 

The  polynomial  multiplicand,  (x  +-  y),  is  multiplied  by  each 
separate  term  of  the  polynomial,  (a  +  6),  and  the  resulting 
products  are  added.  The  process  will  be  clearly  understood 
from  the  following  comparison  : 

Numerical  Illustration :  Algebraic  Illustration : 

a  +    5  Multiplicand. 

a  +    7  Multiplier. 

a2+    ha 

+    7  q  +  35 
a2  +  12  a  +  35    Product. 

Explanation : 
a  (a  +  5)  =  a2  +  5  a 
7  (a  +  5)  =  7  a  +  35 
a2  +  (5  a  +  7  a)  +  35  =  a2  +  12  a  +  35 

We  have,  therefore,  the  following  general  process  for  multi- 
plying a  polynomial  by  a  polynomial : 

69.  Arrange  the  terms  of  each  polynomial  according  to  the 
ascending  order  or  the  descending  order  of  the  same  letter. 

Multiply  all  the  terms  of  the  midtiplicand  by  each  term  of  the 
multiplier.     Add  the  partial  products  thus  formed. 


12  10+2 

13  10+    3 
36      100  +  20 

12                  30  +  6 
156  =  100  +  50  +  6 

Explanation : 
0  (10  +  2)  =  100  +  20 
3  (10  +  2)  =    30  +    6 
00  +  (20  +  30)  +  6  =  156 

46  MULTIPLICATION 

Illustrations : 

1.  Multiply  2  a  +-  7  by  3a- 8. 

2(1     +       7  T,         , 

3      _    8  Explanation : 

6  a2  +  21  a  3  a  (2  a  +  7)  =  6  a2  +  21  a 

-  16  a  -  56  -  8  (2  a  +  7)  =  -  16  a  -  56 

6a2  +   sa_56  Result.       6a2+ (21a-16a) -56  =  6a2  +  5a-56. 

2.  Multiply  a3-2a2  +  3a-2  by  a2  +  3a-2. 

a3_2a2+3a_2  Explanation : 

a2+3q-2 a2  (a3-2  a2+3  a-2)=a6-2  a4+3  a8-2a2 

a5_2a4+3a8-  2  a2  3  a  (a3-2  a2+3  a-2)=3  a4-6  a8+9a2-6a 

+3a4-6a8+  9  a2-  6  a    -2  (a3-2  a2+3  a-2)  =  -2  a3+4a2-6a+4. 

—2  a  +  4a  —  6  a+4      Adding  the  partial  products,  we  have 

a6+    a4-5a8+lla2-12a+4         a5  + a*  -  5a8  + 11  a2  -  12a  +  4. 

Expressions  given  with  their  terms  not  arranged  should 
both  be  arranged  in  the  same  order  before  multiplication. 

3.  Multiply  l  —  7x*  +  a?  +  5x  by  -4z-  1  +  23J2. 

Arrange  both  multiplicand  and  multiplier  in  the  descending  order. 
x8  —  7se2  +  5se-}-l       (Let    the    student  complete   this  multiplication, 

2x  —  4a:  —  1 writing  out  a  complete  explanation  in  the  same 

form  as  those  accompanying  examples  1  and  2.) 

70.  Checking.  A  convenient  check  for  work  in  multipli- 
cation can  usually  be  made  by  the  substitution  of  a  small  num- 
ber as  shown  in  addition.     Thus,  in  Ex.  2,  if  a  =  2 : 

Multiplicand  a3-2a2+3a-2=  8-8  +  6-2=4 

Multiplier  a2  +    3a  -2=  4+6-2=_8 

Product  a5  +  a4  -  5  a3  +  11  a2  -  12  a  +  4  =  32  +  16  -  40  +  44  -24  +4=32 

It  is  well  to  remember  that  this  check  will  not  always  serve 
as  a  test  for  both  coefficients  and  exponents.  If  the  value,  1, 
had  been  used  above,  only  the  coefficients  would  have  been 
tested,  for  any  power  of  1  is  1. 


MULTIPLICATION  OF  MISCELLANEOUS  TYPES  47 

Exercise  9 

Multiply : 

1.  4  a +  7  by  3  a  +  5.  7.  7  ac-  3  by  5  ac  4-1. 

2.  3a  +  4  by  7<c-3.  8.  llafc-3  by  5a6c  +  2. 

3.  5ra-9  by  4m  +  7.  9.  4^-7  by  3a^  +  l. 

4.  2  c  — 5  y  by  3  c  + 11 2/-  10.  7  mw  —  a?3  by  3  raw  +  2  x3. 

5.  4  ra  — 11  by  4  ra  +  11.  11.  16  x3  -  10  xy  by  5  ar*  + 11  a^. 

6.  5cd-7  by  6cd  +  5.  12.  <lab2-b3xy  by  ab2-3b3xy. 

13.  c3 -3^  +  3 c-1  by  c2- 2c  +  1.  .    . 

14.  ra2-2ra+l  by  ra2-2ra  +  l. 

15.  a3  +  2a2-3a  +  4  by  a2-3a-l. 

16.  3d2-5d  +  2  by  d2-d-l. 

17.  4/-7?/2-3y  +  2  by  ?/2-52/-4. 

18.  x3  —  x?y  +  xy2  —  y3  by  a2 -f  cci/ +  2/2- 

19.  27  -  18  ra  +  12  ra2-  8  ra3  by  3  +  2  ra. 

20.  1  -  2  ac  +  4  aV  -  8  a3c3  4- 16  a4c4  by  1  4-  2  ac. 
2i.  a;-7-3ic24-»3  by  a-3  +  2aA 

22.  ra2-2ra4  +  7-2ra3-ra  by  2  ra2- 9 -3  ra3-2  ra. 

23.  12-7x  +  5x3-2x2  by  -Saf  +  x-Sx^x3. 

24.  -11  a-  7  a3 4-17  +  a4 -3a2  by  3a- 10- 7a2 4- 2 a3. 

MULTIPLICATION  OF  MISCELLANEOUS  TYPES 

71.    Illustrations : 
1.   Multiply  a 4- &  4- 2  by  a 4-^—2. 
a  +      6  +  2 
a  +       6-2 
a2  +     a6  +  2  a 

+     ab  +  62  +  2  6 

-2a  -26-4 


a2  +  2  a6  +  62  -  4    Result. 


48  MULTIPLICATION 

2.  Multiply  a2  +  62  +  c2  +  2  ab  —  ac-  be  by  a  +  b+c. 

Arranging  in  the  descending  powers  of  a  : 
a?  +  2ab  -  ac+      b2  -      bc  +    c2 

a  +      b  +     c 

a8  +  2  a2b  -  a2c  +     ab'2  -    abc  +  ac2 

+    a2&  +2  a&2  -    a&c  +  68  -  b2c  +  6c2 

+  a?c +2  abc-  ac2         +  b2c  -  6c2  +  c8 

a8  +  3  a26  +3  a&2  +  68  +  c8    Result 

3.  Multiply  (a  -  2)3. 

(a  -  2)8  =  (a  -  2)2(a  -  2) 

=  (a2-4a  +  4)(a-2) 

=  a8  -  6  a2  +  12  a  -  8.     Result. 

4.  (a  +  5)(a  +  6)(a-3). 

(a  +  6) (a  +  6) (a  -  3)  =  [(a  +6) (a  +  6)] (a  -  3) 

3=  (a2+lla  +  30)(a-3) 


5. 

(a  +  x)(a  +  y). 

a  +  se 

a2  +  a£ 
+  ay  +  xy 

Or, 

a2  +  ax  +  ay  +  xy 
a2  +  (a;  +  y)a  +  icy.    Result. 

a8  +  8  a2  -  3  a  -  90.     Result. 

6.    (a-x)(a-y). 

a  —  x 
a-y 
a2  — ax 
-ay  +  xy 

a2  — ax  — ay  +  xy 

Or,  a2  —  (as  +  y)a  +  Xf.     Result. 

Exercise   10 

Perform  the  following  indicated  operations : 

1.  (c  +  x  +  3)(c  +  x-3). 

2.  (a  +  m  +  y)(a  +  m  —  y). 

3.  (a  +  c  +  m  +  x)(a  +  c  —  m  —  x). 

4.  (ra24-  2mn-\-n2 +  y)(m?  +  2mw  +  w2-  ?/). 

5.  (c2  +  aj2-r-z2-r-2ca;  — cz— #z)(c  +  #  +  z). 

6.  (m  +  a;  +  w  +  2/)(m  —  w  —  #  —  y). 


MULTIPLICATION   OF  MISCELLANEOUS  TYPES  49 

7.  (a  +  3 bx)2.  9.    (3ra-2?i2)3.       11.    (5 a2 -2 ay)*. 

8.  (3  ran- 2  n?/)2.    10.    (4  a2 -7  or5)2.       12.   (3  cd2  -  5  cdxf. 

13.  (C2_C_1)(C2  +  C_1)(C2_1)> 

14.  02-n-f  l)(n2  +  n  +  l)(n4-»2  +  l). 

15.  (x  +  6)(x-7)(x-3). 

16.  (2x-5)(3x  +  l)(2x  +  5)(3x-l). 

17.  (cd-3)(cc*  +  7)(2cd-l)(3cd  +  2). 

18.  (a2-l)(a2-5)(a2  +  l)(a2  +  5). 

19.  (9x2-3x  +  l)(±x2  +  2x  +  T)(3x  +  l)(2x-l). 

20.  (a  +  &)(a  +  m).  24.  (a  +  c)(&  +  d). 

21.  (3a  +  x)(2a  +  y).  25.  (3a  +  2&)(2c-5d). 

22.  (am  —  x)(am  —  y).  26.  (m3  +  2)(m2  — m). 

23.  (3cd-ra)(2cd  +  n).  27.  (m  +  w+l)(m-y). 

Perform  the  indicated  operations  and  simplify  : 

28.  5x2-(x  +  l)(2x-3)-3x(x-l). 

29.  (a-2)2+(a  +  3)2-2(a2  +  a  +  4)-l. 

30.  (2m-l)(m  +  3)-(4m  +  l)(2m-5)-(l-3m)(l+2m). 

31.  (2a-3)2-3a(a-2)-(3-a)2. 

32.  cd(cd  + 1)  +  cd(cd  +  l)(od  +  2)  -  <?d\cd  +  4). 

33.  b2(b2  +  6  -  1)  -  6(62  -  6  +  1)  -  &2(&2  -  1). 

34.  mn(mn  —  1)  —  [(ran  —  l)2  —  (1  —  ran)]. 

35.  (m-2n-3)2  +  ra(2n  +  3-ra)  +  2n(ra-2n-3). 

36.  (l-x)(l-y)+x(l-y)+y. 

37.  a(b  —  ra)  +  b(m  —  a)  +  ra(a  —  6). 

38.  (x  +  a)(x  —  a)  +  (a  4-  2) (a  —  z)  +  (x  +  z)(z  —  a). 

39.  a\b  -x)  +  b\x  -  a)  +  x*(a  -  b )  +  (b  -  x) (x  -  a)(a  -  b). 

40.  (a  +  &  +  c  +  d)2  —  (a  —  b  —  c  —  d)2. 

41.  (a  +  x  +  l)2  +  2(a+x  +  l)(a  +  x  —  l)  +  (a  +  x  —  l)2. 

42.  (a+lf_3(a  +  l)2(a-l)+3(a  +  l)(a-l)2-(a-l)3. 

SOM.    EL.    ALG.  4 


50  MULTIPLICATION 

72.  Multiplication  with  Literal  Exponents.  The  literal  expo- 
nent is  constantly  used  in  the  later  discussions  of  algebra, 
and  familiarity  with  this  form  is  readily  attained  in  the  pro- 
cesses of  multiplication  and  division. 

Illustrations : 

1.   Multiply  a3w4-a2w-2am  +  3by  a^  +  cr  — 1. 

aBm  +  a2tn      _  2  am  +  3 
a2m  +  am        _  1 

abm  _j_  aim     _  2  aSm  +  3  a2m 

4  a4m    4  a3m    -  2  a2m  +  3  am 

-  a3n»    -  a2m    4  2am-S 


abm  +  2  a4m  -  2  a3™  +5  am  -  3    Result. 

2.   Multiply  an+1  -  2  of  +  3  a*"1  -  xn~2  by  a;"  -  3  a;""1. 

»»+1  -2sen  +  3  xn  -1  -  xn~2 
xn      —  3  x"-1 

£2n+l  _  2  X2n  +  3  X2n_1  —        x2n~2 

-3x2n  +  6  a:2"-1  -   9  x2n~2  4  3  a2"-8 
X2n+i  _  5  x2n  +  9  x2"-1  -  10  x2"-2  4  3  x2"-8    Result. 

Exercise  11 

Multiply : 

1.  aw  +  7by  aw  +  3.  4.   2  am+1  -  7  by  3  aw+1  -  4. 

2.  am  —  4  by  aw  —  9.  5.3  a""1  —  11  by  5  aw_1  —  8. 

3.  an+7by  xn-S.  6.   4  aw+2-7  a  by  3  am+2  +  2  a. 

7.  a;n  +  2a;n-1-  3  a;"-2  —  xw~3  by  x  +  1. 

8.  am+s  —  am+2  —  am+1  +  am  by  am+1  4-  a"1. 

9.  a4"  —  ar*w  +  a;2n  —  xn  + 1  by  af  + 1. 

10.  cTO  —  cm~l  4-  cm~2  -f-  cw~3  by  c2  —  c. 

11.  a8m  -  3  a6w  -f  3  a4m-a2m  by  a2w  +  l. 

12.  an+1  —  2  aw  —  3  a;71-1  +  4  of1-2  by  a*  —  3  x"'1. 

13.  a8"1  —  a2m6n  4-  amb2n  —  b3n  by  a2m  +  ambn  +  &2w. 

14.  2  a;m+3n+1  +  a;2m+2n  +  3  iC3w»+w-1  4.  aj4m_2  +  x5m~n~s 

by  ajw+n  -1  -f-  cc2**-2  4-  ar5"1-"-3. 


MULTIPLICATION   OF   MISCELLANEOUS   TYPES  51 

73.  Multiplication  with  Detached  Coefficients.  In  many  cases 
the  labor  of  multiplication  and  division  is  lessened  by  the  use 
of  detached  coefficients. 

By  Ordinary  Multiplication  :  By  Detached  Coefficients : 

2x3-3x2+4x-5  2-3+4-5 

3x2  +  2x  -    1  3  +  2-1 


6  xb  -  9  x*  +  12  x3  -  15  x2  6-9  +  12-15 

+  4x4-   6x3+   8x2-10x  _l_4_6+8-10 
-   2x3+   3x2-   4x  +  5  _   2+   3-   4  +  5 

6^-5 x4  +   4x3-   4x2-14x  +  5  6-5+   4-   4-14  +  5 

* 

Since  the  product  of  two  expressions  is  an  expression  whose  degree  is 

the  sum  of  the  degrees  of  the  given  expressions  (Art.  67),  we  supply  the 

necessary  x-factors  for  the  coefficients 

6      -5      +4      -4       -14     +5, 
obtaining  6  x5  —  5  x4  +  4  x3  —  4  x2  —  14  x  +  5,  the  result. 

Missing  Powers.  If  any  power  of  a  literal  factor  is  miss- 
ing, its  coefficient  is  0,  and  the  term  must  be  provided  for 
in  the  sequence  of  powers  by  an  inserted  0. 

Thus,  (x3-2  x2+  3) (x8+  x-1)  =  (x3  -  2 x2  +0 x  +  3)(x*  +  0 x2  +  x-1). 
Multiplying  with  the  coefficients  detached, 

1-2+0+3 
1+0+1-1 


1 _2+0+3 

+1 _2+0+3 

-1+2-0-3 

1-2+1+0+2+3-3 

Supplying  the  x-factors,  x6-2x5  +  x4+2x2  +  3x-3.    Result. 

For  practice  in  multiplication  with  detached  coefficients,  use 
Ex.  9,  13  to  24  inc. 

At  the  discretion  of  the  teacher  the  foregoing  paragraph 
and  the  corresponding  operation  suggested  under  Division  may 
be  omitted  on  the  first  reading  of  the  text.  For  review  topics, 
however,  the  methods  have  a  distinct  value. 


CHAPTER  V 
DIVISION.    REVIEW 

74.  Division  is  the  process  of  finding  one  of  two  factors  when 
their  product  and  one  of  the  factors  are  given.  The  dividend 
is  the  given  product ;  the  divisor,  the  given  factor ;  arid  the 
quotient,  the  factor  to  be  found. 

THE  NUMBER  PRINCIPLE   OF   DIVISION 

75.  The  Law  of  Distribution.  The  quotient  of  a  polynomial  by 
a  monomial  equals  the  sum  of  the  quotients  obtained  by  dividing 
each  term  of  the  polynomial  by  the  monomial. 

T  i  ax  +  ay  +  az     ax  ,  ay  .  az 

In  general :  —      y    —  =  —  +  —  +  —  =  x  +  y  +  z. 

a  a       a       a 

Numerical  Illustration : 

The  significance  of  this  law  is  that  the  divisor  is  distributed 
as  a  divisor  of  each  term  of  the  dividend: 

The  process  of  division  being  the  inverse  of  the  process  of 
multiplication,  the  laws  governing  signs,  coefficients,  and  ex- 
ponents in  multiplication  form,  when  inverted,  the  correspond- 
ing laws  for  division. 

SIGNS  IN  DIVISION 

By  Art.  56:  (+a)(  +  6)=+a&       Hence,  +a&^+a=+&  (1) 

(— a)(—  b)=+ab     inversely,  -\-ab-. —  a=—  b  (2) 

ByArt.57:  (  +  a)(-6)=-a&  by  -ab+4-a=-b  (3) 

(— a)(-h&)  =  —  ab  division:  —  ab-. —  a=+b  (4) 
52 


COEFFICIENTS  AND  EXPONENTS  IN  DIVISION         53 

Therefore,  from  (1)  and  (4),  and  from  (2)  and  (3)  : 

76.  Like  signs  in  division  give  a  positive  result. 

77.  Unlike  signs  in  division  give  a  negative  result. 

COEFFICIENTS  IN  DIVISION 

78.  The  coefficient  of  a  quotient  of  two  algebraic  expressions 
is  obtained  by  dividing  the  coefficient  of  the  dividend  by  the 
coefficient  of  the  divisor. 

That  is,  12  a  -=-  3=  (12  --  3)  a  =  4  a. 

24  mx-=-  8  =  (24  w  -=-  8)  x  =  3  mx. 

EXPONENTS  IN  DIVISION 

By  definition,  a5=axaxaxaxa, 

a3  =  a  x  a  x  a. 

Therefore,  *  =  axaxaxaxa  =a*. 

a3  a  x  a  x  a 

That  is,  a5  +  a3  =  a5"3  =  a2. 

In  general,  therefore,  we  have  the  following : 
If  m  and  n  are  any  positive  integers  and  m  is  greater  than  n  : 
am=  a  x  a  x  a  •••  to  m  factors, 

fln  =  axflXfl-ton  factors. 
am     a  x  a  x  a  •••  to  m  factors 


Therefore, 


an     a  x  a  x  a  •••  to  n  factors 
—  a  x  a  x  a,'"  to  m  —  n  factors 


Hence,  the  statement  of  the  second  index  law  is  made  as 
follows : 

79.  The  quotient  of  two  powers  of  a  given  factor  is  a  power 
whose  exponent  is  the  exponent  of  the  dividend  minus  the  exponent 
of  the  divisor. 


54  DIVISION.    REVIEW 

80.  Any  quantity  with  a  zero  exponent  equals  1. 

By  Art.  79,     —  =  am~m  =  a0.     But,  —  =  1.     Therefore,  a0  =  1. 

am  am    ■ 

This  principle  inconstantly,  in  use  in  division.     For  example, 

a%8 
We  would  obtain  the  same  result  by  the  old  method  of  saying  "xB  in 
Xs  once"  ;  and  the  quotient  "  1 "  obtained  in  this  way  is  a  factor  of  the 
complete  quotient. 

DIVISION  OF  A  MONOMIAL  BY  A  MONOMIAL 

By  application  of  the  laws  established  for  signs,  coefficients, 
and  exponents,  we  obtain  a  process  for  the  division  of  a  mono- 
mial by  a  monomial. 

Illustration : 
1.   Divide  -  35  aVy4  by  7  aWy2. 

-  35  aHy  =  _  5  ^-8358-8^4-2  -  _  5  a2y2>     Result. 
7  a?xsy2 

Hence,  to  divide  a  monomial  by  a  monomial: 

81.  Divide  the  coefficient  of  the  dividend  by  the  coefficient  of 
the  divisor,  annexing  to  the  result  the  literal  factors,  each  with  an 
exponent  equal  to  its  exponent  in  the  dividend  minus  its  exponent 
in  the  divisor. 

Oral  Drill 

Give  orally  the  quotients  of : 

1.  2.  3.  4.  5. 

a3)a[_  a*)- a*  -x2)^         -m^-m7  -ab)-a2b3 

6.  7.  8.  9. 

4q)-12a8      -Sx^Wx'y     11  xy)-U&y*     - 7  a8b) -  28  asb2 

10.  11.  12.  13.  14. 

-  36  my        -  39  c3dx        42  afyV        -  108  ah1        -  84  asbc2d 
9  m2y'2  -13c2x  -7xfz*        -  IS  a2z         - 12  a2bcd 


DIVISION  OF  POLYNOMIAL  BY  MONOMIAL  55 

x      DIVISION  OF  A  POLYNOMIAL  BY  A  MONOMIAL 

Applying  the  Law  of  Distribution  (Art.  75),  and  the  prin- 
ciples governing  the  division  of  monomials  (Art.  81),  we  obtain 
a  process  of  division  when  the  dividend  is  a  polynomial  and 
the  divisor  a  monomial. 

By  Art.  75:  ax  +  &  ±  az  =  ^  +  &  +  ^  =  x  +  y+  z. 

a  a       a       a 

In  practice  the  usual  form  of  expression  is    a)ax  +  aV  +  az 

x  +  y  +  z         Quotient. 

Illustration : 

1.   Divide  —  6  mAn  -f  10  m3n2  — 14  m2n3  +  4  ran4  by  —  2  mn. 

-  2  mn)  -  6  m4n  +  10  w3w2  -  14  m2n3  +  4  wn4 

3  w3    -    5  m%    +    7  mn2  -2n3       Kesult. 
Each  term  of  the  quotient  is  obtained  by  applying  the  principle  of 
Art.  81,  for  in  the  division  of  each  term  by  the  monomial  divisor  we  have 
a  division  of  a  monomial  by  a  monomial. 

Hence,  to  divide  a  polynomial  by  a  monomial : 

82.  Divide  separately  each  term  of  the  polynomial  by  the 
monomial  and  connect  the  terms  of  the  resulting  polynomial  with 
the  proper  signs. 

Exercise  12 

Obtain  the  quotient  of : 

1.  2.  3. 

x)x-7xy  4  m)12m2-16m8  -  2  q)8  a3b  - 10  ac 

4.  5. 

-  9  ac)18ac-27a3c3  aV-4a3  +  2a2 

6.  7. 

-  4  m)  8  m4  - 1 2  m3  +  8  m2  -  4  m      -  3  a2xz)  -  9  a3x V  -  6  a  W 

8.  14  x3y  —  21xy3by  7  xy. 

9.  4c3d-2cd  +  6cd3by  -2cd 


56  DIVISION.     REVIEW 

10.  -  25  a?x2  +  20  a2x2  - 15  aV  by  -  5  a¥. 

11.  5  a36  - 15  a2b2  +  30  ab3  by  5  ao. 

12.  -36m4-48m5H-60m6-72m7  +  84m9by  -12m3. 

DIVISION  OF  A  POLYNOMIAL  BY  A  POLYNOMIAL 

By  Art.  69 :  (a  +  6)  (z  +  y)  =  ax  +  ay  +  6x  +  6y. 
Then,  by  definition,  ax  +  ay  +  6x  +  fry  may  be  considered  a  gri'vera  divi- 
dend, and  x  +  y  a  given  divisor.  It  remains  to  determine  the  correspond- 
ing quotient.  Both  dividend  and  divisor  are  first  arranged  in  order,  a 
step  imperative  in  all  algebraic  divisions  with  polynomials.  The  x-term 
is  the  selected  letter  for  the  arrangement.  In  algebraic  divisions  the 
divisor  is  written  at  the  right  of  the  dividend  for  convenience  in  the  later 
steps  of  the  process. 

Dividend  Divisor 

ax  -f-  x  =  a  ax  +  ay  -+  bx  +  by  (%  +  y 

a(x-\-y)=  ax-\-ay+ a  +  b      Quotient,  or  Result. 

bx  -=-  x  =  b  +bx  +  by 

b(x  +  y)=  +bx+by 

The  steps  of  the  operation  are  as  follows : 

(1)  The  expressions  are  arranged  in  the  same  order. 

(2)  The  first  term  of  the  dividend,  ax,  is  divided  by  the 
first  term  of  the  divisor,  x ;  and  the  quotient,  a,  is  written  as 
the  first  term  of  the  quotient. 

(3)  The  divisor,  x  +  y,  is  multiplied  by  the  first  term  of  the 
quotient,  a,  and  the  product,  ax  +  ay,  is  subtracted  from  the 
given  dividend. 

(4)  The  remainder,  bx  +  by,  is  a  new  dividend,  and  the 
process  is  repeated  with  this  new  dividend,  the  divisor  always 
being  the  first  term  of  the  given  divisor. 

The  process  applies  the  law  of  distribution,  for 
ax  +  ay  +  bx  +  by  _  ax  +  ay  .  bx  +  by 
x  +  y  x  +  y         x  +  y 

=  a(x  +  y)      b(x  +  y) 
x+y    +    x+y 


DIVISION  OF  POLYNOMIAL  BY   POLYNOxMIAL  57 

A  simple  comparison  with  a  numerical  process  is  frequently 


an  aid  to  beginners. 

Numerical  Illustration: 

Algebraic  Illustration : 

12)156(13        100  4-  50  +  6(10  +  2 
12                100  +  20          10  +  3 
36                        30  +  6 
36                        30  +  6 

a2  +  5  a  +  6{a  +  2 

a2  +  2  a          a  +  3    Quotient 

3a  +  6 

3a  +  6 

Explanation  :  Explanation : 

100  -r- 10  =  10,  1st  term  of  quo.  a2  -r-  a  =  a,  1st  term  of  quo. 
10(10  +  2)  =  100  +  20.     (Subtract.)      a(a  +  2)  =  a2  +  2  a.     (Subtract.) 

100  +  50  +  6  -  (100  +  20)  =  a2  +  oa  +  6-(a2  +  2c)  = 

30  +  6,  the  new  dividend.  3  a  +  6,  the  new  dividend. 

30  -=- 10  =  3,  2d  term  of  quo.  3  a  +  a  =  3,  2d  term  of  quo. 

3(10  +  2)  =  30  +  6.     (Subtract.)  3(a  +  2)  =  3  a  +  6.    (Subtract.) 

30  +  6  -  (30  +  6)  =  0.  3  a  +  6  -  (3  a  +  6)  =  0. 

Hence,  the  quotient  is  10  +  3.  Hence,  the  quotient  is  a  +  3. 

We  have,  from  these  illustrations  and  from  the  general 
principle,  the  following  process  for  the  division  of  a  polyno- 
mial by  a  polynomial : 

83.  Arrange  both  dividend  and  divisor  in  the  same  order  of 
some  common  letter.  Divide  the  first  term  of  the  dividend  by  the 
first  term  of  the  divisor,  and  write  the  result  as  the  first  term  of 
the  quotient. 

Multiply  the  whole  divisor  by  the  first  term  of  the  quotient  just 
obtained,  and  subtract  the  product  from  the  dividend. 

Regard  the  remainder  as  a  new  dividend  and  proceed  in  the 
same  manner  as  before. 

Illustration :  » 

1.   Divide  a4  +  a3-5  a2  +  13  a- 6  by  a2-2a  +  3. 


Dividend 

a*+    a3 

-  5  a2  +  13  a  - 

-6  (a2 

-2a 

+  3 

Divisor 

a4  -  2  a3 

+  3«2 

a2 

+  3a 

-2 

Quotient 

+  3  a3 

-  8  a2  +  13  a 

+  3a3 

-6a2+    9  a 
-  2 a2  +    4a- 
-2«2+    4a  _ 

-6 
-6 

58  .     DIVISION.    REVIEW 

Explanation : 

(1)  a4  -j-  a2  —  a2,  the  first  term  of  the  quotient. 

(2)  Multiplying  (a2  -  2  a  +  3)  by  a2,  we  obtain  a*  -  2  a8  +  3  a2,  which 
product  is  subtracted  from  the  given  dividend.  The  remainder,  which 
must  have  three  terms,  is  3  az  —  8  a2  +  13  a,  and  this  remainder  is  the 
new  dividend. 

(3)  3  a8  ■+■  a2  =  3  a,  the  second  term  of  the  quotient. 

(4)  Multiplying  (a2  -  2  a  +  3)  by  3  a,  we  obtain  3  a8  -  6  a2  +  9  a, 
which  product  is  subtracted  from  the  last  dividend.  The  remainder, 
with  the  last  term  of  the  given  dividend,  is  — 2a2  +  4a  —  6,  the  new 
dividend. 

(5)  —  2  a2  -*-  a2  =  —  2,  the  third  term  of  the  quotient. 

(6)  Multiplying  (a2-2a  +  3)  by  -  2,  we  obtain  -  2  a2  -f  4  a  -  6, 
which,  subtracted  from  the  last  dividend,  gives  a  remainder  of  0,  com- 
pleting the  division. 

2.   Divide  7x2-xA-xs-\-2x5  +  2-9xhj3x-2x2-2  +  xs. 
Arranging  both  dividend  and  divisor  in  descending  powers 
of  x: 


Dividend 

2x5-    a*- 

X*  + 

7s2- 

-9s  +  2(a;8- 

-  2  a2  +  3  a  -  2 

Divisor 

2x5-4xi  +  6x8- 

4x2 

2  a5 

+  3a;  —  1 

Quotient 

+  3**- 

-7z3  +  llx2- 

-9x 

+  3z*- 

-6z8  + 

9x2- 

-6x 

■    SC3  + 

2x2- 

-3x  +  2 

.  cc8  + 

2z2- 

-3x  +  2 

In  certain  expressions  the  intermediate  powers  of  the  terms 
of  the  dividend  appear  during  the  process.  In  such  cases  order 
must  be  observed  in  subtractions. 

3.   Divide  x3  —  y3  by  x  —  y.         4.   Divide  a4  — 16  by  a  +  2. 


X*  _      yS  (X-y 

a*             -16(a  +  2 

x*-x2y*  x2  +  xy  +  y2 

a*  +  2  a3           a8  -  2  a2  +  4  a  - 

-8 

+  x2y* 

-2  a8 

+  x2y  -  xy2 

-  2  a8  -  4  a2 

+  xy2  -  y* 

+  4  a2 

+  xy2  -  y* 

+  4  a2  +  8  a 

-  8  a  -  16 
-8a -16 

DIVISION  OF  POLYNOMIAL  BY  POLYNOMIAL  59 

5.   Divide  a3-3  a&c+-63  +  c3  by  a  +  6  +  c. 

a3  -  3  abc  +  63   +  c3  (q  +  &+c' 

a3  +     d2b  +  a2c         (a2  -  ab  -  ac  +  62  -  be  +  c2 

-  a26  -  a2c  -  3  a5c  +  63  +  c3 

-  <z25  -  ab2  -    abc 

-  a2c  +    aft2  -  2  abc  +  63  +  c8 

—  a2c  —    abc  —  ac2 


+    ab'2  -    abc  +  ac2  +  b3  +  c* 
+    ab2 +  63  +  b2c 

—  abc  +  ac2  —  b2c  +  c8 

-  abc -  b2c-  be2 

+  ac2  +  be2  +  c8 

+  ac2 +  6c2  +  c8 

Note  that  in  all  new  dividends  the  letter  a  is  selected  for  arrangement. 

84.   Checking.     The   work  in    division   may   sometimes   be 
checked  as  shown  in  multiplication.     Thus,  in  Ex.  1,  if  a  =  2 : 

Dividend     a4  +  a3  -  5  a2  +  13  a  -  6  _  16  +  8-20+26-6  _  24  _  g 
Divisor  a2  -  2  a  +  3  4-4  +  3  3 

Quotient     a2  +  3a-2  =  4  +  6-2  =  8. 

It  is  well  to  remember  that  this  method  of  testing  is  not 
always  reliable. 

Exercise  13 

Divide :  » 

1.  2o2  +  7a>+-6  by  x  +  2. 

2.  2c2  +  5c-12by  c  +  4. 

3.  3m2+-llm-4by  3m-l. 

4.  3z2  +  14z+-15by*  +  3. 

5.  16  -  8  a  +-  a2  by  4  -  a. 

6.  72 +  m-m2  by  8  +  m. 

7.  7a262  +  123a6-54by7a6-3. 

8.  20a2-47a6  +  2162by  4a-76. 

9.  21  c4d4  -f  36  c2dV  + 15  a4  by  3  c2cP  +  3  jb2. 

10.   10  a664  +  23  a362cd?  -  21  c2^  by  10  a?b2  -  7  cd2. 


60  DIVISION.    REVIEW 

11.  a3+3a2  +  3a  +  lby  a  +  1. 

12.  c3^-\-c2x2-3cx-6  by  cx-2. 

13.  15  —  8  mn  +  6  m2n2  —  m37i3  by  5  —  mn. 

14.  a4  +  2  a36  4-  2  a262  +  ab3-  6  64  by  a?  +  ab-  2  b\ 

15.  2  ra4n4  —  m3n3x  —  3  m2n¥  +  5  mna?5  —  2  #4 

by  2  m27i2  —  3  mnx  +  2x*. 

16.  8  aW  -  8  aVd  -  4aW  + 11  aW  - 10  acd4  +  3  d5 

by  2a2c2-3aca,  +  a*. 

17.  9  a2- a4- 16  a- a3  + 6 a5  + 3  by  5  a -2  a2- 1  +  3  a3. 

18.  16a;6  +  l-4a2-4a;4by  2  a2- 1  +  4  ^-2  a. 

19.  5m2-6-llm4  +  6m6-5m8  +  2ra10by  2  +  2m4-m2. 
20.5  mV  - 13  m6?i  +  6  m3n4  +  3  m¥  +  6mn6  +  6m7 

4- 8 ti7  — 15 ra47i3  by  3 m4  —  mti3  —  2 w3ti -2n4  +  m2n2. 

21.  m2  —  w2  by  m  -  w.  24.   ra4  —  ti4  by  ra  +  ti. 

22.  m2-7i2by  ra  +  7i.  25.   8  a3-27  by  2 a-3. 

23.  m3  —  n3  by  m  —  n.  26.   m6  —  32  n5  by  m  —  2  w. 

27.  81 -w4  by  3 +%. 

28.  27  +  a6by  3  +  x2. 

29.  a^-27^by  a;-32/2. 

30.  27 a,-3  +  64 y5  by  3a  +  4?/2. 

31.  16ra4n4-81aV6by  2m7i  +  3cey. 

32.  m2  +  2  mn  +  ti2  —  x2  by  m  +  n  +  x. 

33.  c2  — 4c  +  4  — d3by  c-2  +  d 

34.  m2  +  6m  +  9— 25  a4  by  m  +  S-Bx2. 

35.  16 m2n2  +  40  m7iy  + 25 2/2- 81  by  4ra7i  +  5y  +  9. 

36.  a2  +  m2  +  a?2  +  2  am  +  2  asc  +  2  m#  by  a  +  m  +  aj. 

37.  9m2n2  +  4:x2z2  +  25-12mnxz-30mn  +  20xz 

by  3  m»  —  2  o&  —  5. 

38.  m3  +  7i8+p3  —  3m7ip  by  m  +  n+p. 

39.  a666-2a363  +  lby  a262-2a6  +  l. 


LITERAL  EXPONENTS  AND  DETACHED  COEFFICIENTS      61 

85.  Division  with  Literal  Exponents.     (See  Art.  72.) 
Illustration : 
Divide  8  of1*4  -  18  afn+3  - 13  xm+2  +  9  a^1  +-  2  x™ 

by  4  xm  +  x"1-1  —  2  zm-2. 
8  xm+i  -  18  scm+3  —  13  xm+'2  +  9  xw+1  +  2xm  (4  zOT  +  xm-1  -  2  xm~2 
8xm+4_|_   2xn>+3-   4xm+2 2  x4  -  5  xs  -  x2   Result. 

—  20xw+3—    9xm+2+   9xm+1 

-20x"t+3-    5xm+2+l0xm+1 

—  4xm+2—   xm+1  +  2xm 

—  4  a?n+2  —   xm+1  4-  2  xm 

The  exponent  of  cc  in  the  first  term  of  the  quotient  is  :  (m  +  4)  —  m  = 
m  +  4  —  ra  =  4. 

Exercise  14 
Divide : 

1.  ^  +  3^-18  by  xm-S. 

2.  ar?n-4arJn-20a;n+-3by  xn  +  3. 

3.  Sx4m-SxZm-10x2m-xm  +  lbj  3x^-^30^  —  1. 

4.  x^-S  by  »n-2. 

5.  x^  —  y*"  by  xm  +  2/n. 

6.  16  xm+1  -  46  a;m+2  +  39  xm+3  -  9  af+4  by  8  a;M+1  -  3  aw+2. 

7.  #n+4  +-  an+3  -  4  zn+2  +  5  xn+l  -  3  #n  by  xn+2  +  2  an+1  -  3  af . 

8.  6a^+6-a;^+6+-4a^+6-5a^w+6-a;m+6  — 15  a;6 

by  2  a^m  —  xm  +  3. 

86.  Division  with  Detached  Coefficients.  The  same  process 
described  in  Art.  73  is  of  advantage  in  division,  as  two  simple 
cases  will  illustrate. 

1.   Divide  x*- 3  ar5-  36  x2 -  71  x  -  21  by  x2 - 8  x-  3. 

Detaching  the  coefficients  and  dividing,  we  obtain : 
1  _  3  _  36  _  71  -  21  (1-8-3 
1-8-3  1  +  5  +  7     Coefficients. 

+  5_33_71 

+  5-40-15 

+    7-56-21        z2  +  5  Z  +  7.     Result. 
+    7-56-21 


62  DIVISION.     REVIEW 

If  0  occurs  as  a  coefficient  either  in  the  given  expressions  or 
in  the  process,  the  same  provision  is  made  as  in  multiplication. 

2.   Divide  #5  +  2  a^+^a3- 31  a+-9  ^+-15  by  3-2  x-x2. 

Arranging  the  expressions  in  ascending  powers  of  x  and 
detaching  coefficients,  we  have : 


15 

-31 +  9  +  4+2  +  1  (3- 

-2-1 

15 

-10-5                        5  - 

-7  +  0- 
5-7* 

1    Coefficients. 

-21  +  14  +  4 
-21  +  14  +  7 

+    0-3+2+1 

-3+2+1 

—  Xs.     Result. 

For  practice  in  division  with  detached  coefficients  use  Ex.  13, 
11  to  20,  inc. 

GENERAL  REVIEW 

Exercise  15 

1.  Show  that  (x-l)2-(x-2)=l  +  (x-l)(x-2). 

2.  Simplify(a+l)2-(a+-l)(a-l)-[a(2-a)-(2a-l)]. 

3.  Prove  that  (a  +  m)(a  —  m)  +-  (m  -f- 1) (m  —  1) 

+  (l-a)(l+a)  =  0. 

4.  Divide4a)6-2^-a?-2^-2a;-lby2^-£2-a;-l. 

5.  Add   the   quotient    of    (a3  —  1)  -f-  (x  —  1)    to    that    of 
(tf-2x  +  l)  +  (x-l). 

6.  What  is  the  coefficient  of  ac  in  the  simplified  form  of 
(ac+-3)2-3ac(ac-l)? 

7.  Simplify  (m  +  l)(m  +  3)(m  +  5)  -(m-l)(m-3)(ra-5). 

8.  Show  that  (x  +-  y  +-  z  —  1)  (x  +-  y  —  z  +- 1)  —  4  (xy  +-  z) 

+  (a-2/+-z  +  l)(l-a;  +  2/  +  z)=:0. 

9.  Divide  2  x2  —  3  xy—5  xz  —  2  y2—  5yz—Sz2  by  2x  +  y  +  z. 

10.    A  certain   product  is  6  a4  +-  4  a3#  —  9  a2y2  —  3  ays  +-  2  ?/4, 
and  the  multiplier  2  a2 -{-2  ay  —  y2.     Find  the  multiplicand. 


11.    Simplify  3[ap  —  2  joj  — 3(2  a?  — 3  aj  +  7){]. 


GENERAL  REVIEW  63 

12.  Show  that  (1  -  3  x  +  x2)2  +  x  (1  -  x) (2  -  x)  (3  -  x)  - 1  =  0 . 

13.  Simplify  2  x2  -3-(3  x  -}-3  x2)-x(x2  -3)-(x  +  1)(2  -  x") 
and  subtract  the  result  from  5  —  2  x. 

14.  Simplify  (a  +  4)  (a  +  3)  (a  +  2)  -  (a  +  3)  (a  +  2)  (a  + 1) 

_(a+2)(a+l)-(a  +  l). 

15.  Prove  that  (1  +  c2)  (1  4-  a)2  -  2  (1  -  be)  (a  -  c) 

=  (l  +  c)2(l+a2). 

16.  Subtract  a +  3  from  the  square  of  a +  2,  and  multiply 
the  result  by  the  quotient  obtained  when  a5  —  1  is  divided  by 
aA  +  az+a2  +  a  +  l. 

17.  Simplify  a  —  x—  [3  a— (x— a)]  +  [(2  x  —  3  a)  —  (a— 2  a)]. 

18.  Divide  c?-3cd+cP  +  l  by  c2- cd -c  +  d2-a'  +  l. 

19.  Find  the  continued  product  of  a2  —  ab  -f  62,  a2  +  a&  -f-  62, 
anda4-a262+64. 

20.  Simplify  (a¥  —  a¥  -f  aa  —  1)  (ax  + 1) 

-  (aV  + 1)  (a»  + 1)  (ax  - 1). 

21.  Multiply  8  a3  —  27  by  a  +  2  and  divide  the  product  by 
2a-3. 

22.  Divide  c  +  36c5-18c2-73c3  +  12  by  _5c  +  4-6c8. 

23.  Simplify  [(x2 +  3x  +  2)(x2  -9)]-*-  [(*  +  3)(^  -«-  6)]. 

24.  Show   that   a^  +  ^+l  —  3a#—  (1  —  ajy)—  ?/(#2 - «)  — 

25.  Divide  82  m4n4  +  40  -  67  mV  +  18  mV  -  45  mV  by 
3  m4n4  —  4  m2n2  +  5. 

26/  What  must  be  the  value  of  m  in  order  that  x2  + 18  x 
+  m  may  be  exactly  divisible  by  x  -f-  4  ? 

27.  Show    that    2(4  +  o^  +  a2  -ax-2a-2x)  =  (2-a)2  + 

(a-2)2  +  (x-a)\ 

28.  What  must  be  the  value  of  m  +  nm  order  that  x*  +  3  Xs 
+  2  x2  +  Tnx  +  n  may  be  exactly  divisible  by  x2  +  2  x  +  1? 

29.  By  how  much  does  (aV  -f-  3  ax  +  2)2  exceed  2(3  a3x3  + 
2a*s*  +  6aa>)? 


CHAPTER  VI 
THE  LINEAR  EQUATION.    THE  PROBLEM 

87.  An  equation  is  a  statement  that  two  numbers  or  two 
expressions  are  equal. 

Thus,  3x  +  5  =  x  +  7. 

88.  The  expression  at  the  left  of  the  sign  of  equality  is  the 
left  member  (or  first  member),  and  the  expression  at  the  right, 
the  riglft  member  (or  second  member)  of  the  equation. 

89.  An  equation  is  a  conditional  equation  if  its  members  are 
equal  for  particular  values  of  the  unknown  quantity. 

Thus,      3x  +  5  =  x  +  7isan  equation  only  when  the  value  of  x  is  1. 

90.  An  equation  is  an  identical  equation  when  i*s  members 
are  equal  for  any  and  all  values  of  the  unknown  quantity. 

Thus,   x2— l  =  (x+l)(x— 1)  is  an  equation  for  any  value  of  x  whatsoever. 

A  conditional  equation  is  usually  referred  to  as  an  equation  ;  an  iden- 
tical equation,  as  an  identity. 

91.  To  solve  an  equation  is  to  obtain  the  value  of  the  un- 
known number  that  will,  .when  substituted  for  that  unknown 
number,  make  the  members  of  the  equation  equal. 

92.  The  value  found  to  make  the  members  of  an  equation 
equal,  or  to  satisfy  the  equation,  is  a  root  of  the  equation.  A 
root  of  an  equation  when  substituted  for  the  unknown  quan- 
tity reduces  the  original  equation  of  an  identity. 

64 


AXIOMS  65 

93.  A  linear  or  simple  equation  is  an  equation  which,  when 
reduced  to  its  simplest  form,  has  no. power  of  the  unknown 
quantity  higher  than  the  first  power.     Thus : 

5  x  =  15  is  a  linear  or  simple  equation  in  x. 
7  y  —  35  is  a  linear  or  simple  equation  in  y. 
3aj  +  2  =  2aj  +  7isa  linear  equation  in  x,  but  is  not  reduced  in  form. 
(x  +  5)2  =#2  +  7a;  +  6isa  linear  equation  in  x,  for,  when  simplified, 
the  resulting  equation  will  have  only  the  first  power  of  x. 

While  the  final  letters,  x,  y,  and  z,  are  most  commonly  used 
for  representing  unknown  quantities  in  equations,  any  other 
letters  may  and  will  be  used  in  later  practice. 

94.  The   solution  of  equations   is  based  upon   the   truths 

known  as 

» 

AXIOMS 

1.  If  equals  are  added  to  equals,  the  sums  are  equal. 

2.  If  equals  are  subtracted  from  equals,  the  remainders  are 
equal. 

3.  If  equals  are  multiplied  by  equals,  the  products  are  equal. 

4.  If  equals  are  divided  by  equals,  the  quotients  are  equal. 

5.  If  two  quantities  are  equal  to  the  same  quantity,  they  are 
equal  to  each  other. 

In  general,  these  axioms  may  be  illustrated  as  follows. 
Given  the  equation  A  =  B. 

By  Axiom  1  A  =  B  By  Axiom  2  A  =  B 

Add  C,  C=  C  Subtract  C,  C=  C 

A+  C  =  B+  C  A- C  =  B-  C 

Or,  briefly : 

95.  The  same  number  may  be  added  to,  or  subtracted  from, 
both  members  of  an  equation. 

By  Axiom  3  A  =  B  By  Axiom  4  A-B 

Multiply  by  C,       C=  G  Divide  by  C,  C=  C 

AC=BC  4=B 

C     C 

SOM.   EL.   ALG.  — 5 


66  THE  LINEAR  EQUATION.     THE  PROBLEM 

Or,  briefly : 

96.    Both  members  of  an  equation   may  be   multiplied,   or 
divided,  by  the  same  number. 

By  Axiom  5         If  A  =  B  and  B  *=  D  ;  we  have,  A  —  D. 
THE  TRANSPOSITION  OF  TERMS 


97.  Most  equations  are  given  in  such  a  form  that  the  known 
and  the  unknown  terms  occur  together  in  both  members. 
Transposition  is  the  process  of  changing  the  form  of  an  equa- 
tion so  that  the  unknown  terms  shall  all  be  in  one  member, 
usually  the  left,  and  the  known  terms  all  in  the  other.  The 
process  is  based  on  Art.  95. 


Given  the  general  equation,  ax  —    c  =  bx-\-  d 
By  Axiom  1,  adding                          c  =  c 

(1) 
(2) 
(3) 

ax          =bx  +  c  +  d 
By  Axiom  2,  subtracting                 bx  =  bx 

ax  —  bx  —  c  +  d 

■ 

Compare  carefully  (1)  and  (3). 

In  (3)  we  find  c  in  the  right  member  with  its  sign  changed 
from  —  to  +. 

In  (3)  we  find  bx  in  the  left  member  with  its  sign  changed 
from  +  to  — . 

In  general: 

98.  Any  term  in  an  equation  may  be  transposed  from  one 
member  to  the  other  member  if  its  sign  is  changed. 

As  a  direct  consequence  of  the  use  of  the  axioms,  we  have : 

(1)  The  same  term  with  the  same  sign  in  both  members  of  an 
equation  may  be  discarded. 

Given  the  equation,    Sx  +  a  —  n  =  2x-\-a  +  m. 
Whence,  Sx  —  n  =  2  x  +  m. 


THE  TRANSPOSITION  OF  TERMS  67 

(2)  The  sign  y  of  every  term  in  an  equation  may  be  changed 
without  destroying  the  equality. 

Given  the  equation,         —  5x  +  m  =  b  —  a. 
Multiplying  by  —  1,  5x  —  m  =  a  —  b. 

The  sign  of  a  root  in  a  solution  depends  upon  the  law  of 
\  signs  for  division.     Thus : 

5  a:  =  20  5x=-20  -5a  =  20  -5  a;  =-20 

x  =  4:  $e=— 4  x=  —  4  se  =  4 

In  general : 

99.  When  both  members  of  an  equation  are  reduced  to  simplest 
form,  lilce  signs  in  both  members  give  a  positive  root,  unlike  signs 
a  negative  root.  # 

If  the  coefficient  of  the  unknown  quantity  in  a  simplified 
equation  is  not  exactly  contained  in  the  known  quantity,  the 
root  is  a  fraction ;  and  if,  in  a  simplified  equation,  the  member 
containing  the  known  quantities  reduces  to  zero,  the  root  is 
zero. 

That  is:  If    3  a  =  5        -  4  x  =  7  And  if    5x  =  0        -9s  =  0 

x  =  f  x=  -£  x  =  0  x  =  0 

Oral  Drill 

Give  orally  the  roots  of  the  following : 

1.  5x  =  S0.  6.   6z  =  -36.  11.  8aj  =  7. 

2.  7z  =  42.  7.    -7x  =  21.  12.  7y  =  13. 

3.  4a  =  28.  8.    -8x  =  -56.  13.  -5a=16. 

4.  3y  =  -lS.  9.    -3a  =  -39.  14.  7#=0. 

5.  -3y=l$.  10.   52=3.  15.  -3a  =  0. 

THE  VERIFICATION  OF  LINEAR  EQUATIONS 

100.  To  substitute  a  *root  in  an  equation  is  to  replace  the 
unknown  literal  factor  in  each  term  by  the  value  of  the  root 
obtained. 


68      THE  LINEAR  EQUATION.  THE  PROBLEM 

101.  To  verify  a  root  is  to  show  that,  by  the  substitution  of 
this  value,  the  given  equation  reduces  to  an  identity.  The 
verification  of  a  root,  as  illustrated  in  the  solutions  following, 
should  always  be  made  in  the  original  equation. 


THE  GENERAL  SOLUTION  OF  THE  LINEAR  EQUATION 

Illustrations : 

1.   Solve  5x-4:=  3a  +  12. 

5x-4  =  3x  +  12. 

Transposing  3  x  to  the  left  member  and  —  4  to  the  right  member, 

5<c-3x  =  4  +  12. 

Uniting  terms,  2  x  =  16. 

Dividing  both  members  by  the  coefficient  of  x, 

x  —  8,  the  root. 
Verification : 

In  the  original  equation,     5x-4  =  3e  +  12. 
Substitute  8  for  as,  5  (8)  -  4  =  3  (8)  +  12. 

40  -  4  =  24  +  12. 
36  =  36. 

Therefore,  8  is  the  correct  value  of  the  root,  for,  by  substituting  8  for 
x  in  the  original  equation,  we  obtain  an  identity. 


2.   Solve  5«-[3-(ss-2a>-l)J  =  -  10. 


5  x  -  [3  -  (x  -  2  x  ~  1)]  =  -  10. 
Removing  parentheses,  5  x  —  3  +  x  —  2  x  +  1  =  -  10. 

Transposing,  &x  +  x  —  2x  =  +  S  —  1  —  10. 

Uniting,  4  x  =  —  8. 

Dividing  both  members  by  4,  x  =  —  2,  the  root. 

Verification : 


In  the  original  equation,  5  x  -  [3  -- (x  —  2  x  -  1)]  =  -  10. 

Substitute  -  2  for  x,  5( -2)  -  [3  -  ( -  2  -  2(-2)-l)]  =  -  10. 
Simplifying,  *    -  10  -  3  -  2  +  4  +  1  =  -  10. 

-  10  =  -  10. 


GENERAL  SOLUTION  OE  THE  LINEAR  EQUATION        69 

3.    Solve  (x  +  3)(2x-5)  =  2(x-2)2-2(x  +  l). 

(x  +  3)  (2  x  -  5)  =  2  (x  -  2)2  -  2  (as  +  1). 

Multiplying,  2  x2  +  X  -  15  =  2  (x2  -  4  x  +  4)  -  (2  x  +  2). 

Removing  parentheses,    2x2  +  x  —  15  =  2 x2  —  8x  +  8  —  2x  —  2. 

Discarding  x2-terms  and  transposing, 

x  +  8x  +  2x  =  15  +  8-2. 

Uniting,  11  x  =  21. 

x  =  ff ,  the  root. 
Verification : 

Substituting  \ \  for  x  in  the  original  equation, 

(H  +  3)(£i-5)=2(fi-2)2-2(fi  +  l). 
(ff)(-i!)=2(-T1T)2-2(!f). 
-m  =  tIt  -  Hf 

4.'  Solve  .3  25  -  (x  -  3)2  =  3.25  -  (x  -  3.5)  (a  +  2.1). 

.3  x  -  (x  -  3)2  =  3.25  -  (x  -  3.5)  (x  +  2.1). 
Multiplying,  .3  x  -  (x2  -  6  x  +  9)  =  3.25  -  (x2  -  1.4  x  -  7.35). 
Removing  (  ),  .3  x  -  x2  +  6  x  -  9  =  3.25  -  x2  +  1.4  x  +  7.35. 
Transposing,       .3  x  +  6  x  -  1.4  x  =  9  +  3.25  +  7.35. 
Uniting,  4.9  x  =  19.6. 

Dividing  by  4.9,  x  =s  4,  the  root. 

Verification : 
Substituting  4  for  x  in  the  original  equation, 

.3  (4)  -  (4  -  3)2  =  3.25  -  (4  -  3.5)(4  +  2.1). 
1.2  -  1  =  3.25  -  (.5)  (6.1). 
.2  =  3.25  -  3.05. 
.2  =  .2. 

From  the  foregoing  we  may  state  the  general  method  for 
the  solution  of  linear  equations: 

102.  Perform  all  indicated  multiplications  and  remove  all 
parentheses. 

Transpose  the  terms  containing  the  unknown  quantity  to  one 
member y  and  all  known  terms  to  the  other  member  of  the  equation. 

Collect  the  terms  in  each  member. 

Divide  both  members  by  the  coefficient  of  the  unknown  quantity. 


70  THE  LINEAR  EQUATION.     THE  PROBLEM 

Exercise  16 

Find  and  verify  the  roots  of  the  following : 

1.  4a;  +  5  =  3aj  +  9.  6.    7  -2x  =  9  x  —  92. 

2.  6  x  -j-  7  =  5  x—  11.  7.    5a  —  l  =  3x  +  l. 

3.  7x-2-5x-12  =  0.  8.    3#  +  7-a=-3. 

4.  8a-12  =  3a;  +  8.  9.    -  7  a-  5  =  -  5  +  2x. 

5.  4«  +  5  =  x- 28.  10.    12a>-7  — 14a;  =  12a>. 

11.  5a>-  (2  a +  3)  =  12. 

12.  (2a-l)-15  =  4-(5-7a). 

13.  13-(2a>  +  ll)  =  6-(«  +  l). 

14.  3«-2-(l-3a>)  =  0. 

15.  12-2(a>-5)  +  [3aj-(2-aj)]  =  l. 

16.  ll-(2«-3)--5=-3»-(5-7a>). 

17.  (x+l)(x  +  3)  =  (x-2)(x-B). 

18.  (2a-3)(x-7)  =  (a?-l)(a  +  4)  +  a;2. 

19.  (aj-3)2  +  2(a-4)2-3(a-5)(a;  +  5)  =  7. 

20.  5(a-l)2-3(a-2)2==(2a-l)(3  +  a;)-6. 

21.  4[3«-2(aj8  +  l)]  =  7-4s(2a>-16). 

22.  -  [2  (x  -  3)(a>  -  5)  -  (a  +  7)(3  -»)]  =  -  3  (a2+  3). 

23.  (a,-  +  2)3-(x-l)3-(3a;  +  l)(3a;-4)=0. 

24.  2.7  x  -  (11  - 1.3  x)  -  6.7  x  =  .62  +  .4  a  -  11. 

25.  0.007  x  -  2  (.0035  g  +  .07)  =  .017-  (.14  -  .85  x). 

THE  SOLUTION  OP  PROBLEMS 

103.   A  problem  is  a  question  to  be  solved. 

In  general,  a  problem  is  a  statement  of  conditions  involving 
an  unknown  number  or  numbers.  We  seek  the  value  of  that 
unknown  number,  and  by  assuming  a  literal  symbol  for  the 
unknown  we  are  able  to  state  the  given  conditions  in  terms  of 
that  unknown. 


LITERAL   SYMBOLS  FOR   UNKNOWN   QUANTITIES        71 

104.  The  solution  of  a  problem  is  (1)  a  translation  of  the 
language  into  the  symbol-expression  of  algebra  by  means  of 
an  assumed  value  for  the  unknown ;  (2)  the  translation  of  the 
given  conditions  into  equations ;  and  (3)  the  finding  of  the 
root  of  the  derived  equation.  The  result  of  a  solution  should 
always  be  verified  in  the  given  conditions. 


LITERAL  SYMBOLS  FOR  UNKNOWN  QUANTITIES 

105.  A  few  simple  statements  will  illustrate  the  ease  with 
which  the  conditions  of  a  problem  are  translated,  or  expressed, 
in  algebraic  symbols.  It  will  be  seen  that  in  each  case  we 
first  assume  a  value  for  an  unknown  quantity  upon  which  the 
statements  seem  to  depend,  and  then  write  expressions  for  the 
different  statements. 

Illustrations : 

1.  If  a  denotes  a  certain  number,  write  an  expression  for  10 
more  than  x. 

Since  we  may  assume    x  =  the  given  number, 
By  adding  10,        x  +  10  =  the  required  number. 

And,  fulfilling  the  given  condition,  we  have  written  a  number  greater 
than  the  given  number  by  10  ;  the  words  "more  than"  being  translated 
by  the  symbol  of  addition,  + . 

2.  In  a  certain  exercise  John  solved  twice  as  many  examples 
as  William.  Write  an  expression  for  the  total  number  of  ex- 
amples both  together  solved. 

We  assume  that       x  =  the  number  that  William  solved. 
Then  2  x  =  the  number  that  John  solved. 

By  addition,  3  x  =  the  number  both  together  solved. 

Here  we  assumed  a  literal  symbol  for  that  particular  number  upon 
which  the  problem  seemed  to  depend ;  that  is,  the  number  of  examples 
solved  by  William.  A  simple  application  of  multiplication  and  addition 
completes  the  translation  of  the  conditions. 


72      THE  LINEAR  EQUATION.  THE  PROBLEM 

3.  A  man  is  y  years  of  age.  (a)  How  old  was  he  10  years 
ago  ?     (b)  How  old  will  he  be  in  z  years  ? 

(a)  Since  y  =  the  number  of  years  in  his  present  age, 
By  subtraction,     y  —  10  =  his  age  (in  years)  10  years  ago. 

And  "years  ago,"  a  decrease,  is  translated  by  the  sign  — . 

(b)  Since  y  =  the  number  of  years  in  his  present  age, 
By  addition,     y  +  z  —  his  age  (in  years)  z  years  from  now. 

And  "  in  "  referring  to  the  future  is  an  increase  translated  by  the 
sign  +. 

4.  m  yards  of  cloth  cost  $  3  per  yard,  and  n  yards  of  silk, 
$  5  per  yard.  Write  an  expression  for  the  total  cost  of  both 
in  dollars. 

At  1 3  per  yard  m  yards  of  cloth  cost  (3  x  m)  dollars  =  3m  dollars. 
At  $  5  per  yard  n  yards  of  silk  cost  (5  x  n)  dollars  =5w  dollars. 
Therefore,  adding,  (3  ra  +  5  ii)  dollars  equals  the  total  cost  of  both. 

5.  Write  three  consecutive  numbers,  the  least  of  the  three 
being  x. 

The  difference  between  any  two  consecutive  numbers  is  1. 
Therefore,  when         x  =  the  first  and  least  number, 

x  +  1  =  the  second  number, 

x  +  2  =  the  third  number. 

6.  Write  three  consecutive  even  numbers,  the  least  being  m. 

The  difference  between  consecutive  even  numbers  is  2. 
Therefore,  as  above, 

m,  m  +  2,  and  m  +  4,  are  the  required  numbers. 

7.  Write  three  consecutive  odd  numbers,  the  greatest  being 
a ;  and  write  an  expression  for  their  sum. 

i    The  difference  between  consecutive  odd  numbers  is  2. 
Therefore,  a  =  the  first  and  greatest  number, 

a  —  2  =  the  second  number, 
a  —  4  =  the  third  number. 
By  addition,     3  a  —  0  =  the  sum  of  the  three  numbers, 


LITERAL   SYMBOLS  FOR  UNKNOWN  QUANTITIES       73 

8.  A  boy  has  x  dimes  and  y  nickels  and  spends  10  cents. 
Write  an  expression  for  the  amount  he  has  remaining. 

Problems  involving  value  must  be  so  stated  that  different  denominations 
are  all  expressed  in  one  and  the  same  denomination. 

Hence,  the  value  of  x  dimes  =  (10  •  x)  cents  =  10  a;  cents. 
Hence,  the  value  of  y  nickels  =  (5  •  y)  cents  =  5  y  cents. 
Therefore,  he  had  at  first,  (10  x  -f  5  y)  cents. 

Subtracting  the  10  cents  he  spent,  we  have  for  an  expression  of  his 
final  amount,  (10  x  +  5  y)  —  10  cents.     Result. 

Oral  Drill  . 

1.  If  x  denotes  a  certain  number,  give  an  expression  for  15 
less  than  x. 

2.  If  y  denotes  a  certain  number,  give  an  expression  for  12 
less  than  the  double  of  y. 

.3.  If  x  denotes  the  number  of  square  inches  in  a  certain 
surface,  how  many  square  inches  are  there  in  m  similar  surfaces  ? 

4.  John  has  x  marbles,  William  y  marbles,  and  Charles  z 
marbles.  What  is  the  expression  for  the  total  number  all 
together  have  ? 

5.  A  boy  had  m  marbles  and  lost  1  of  them.  How  many 
had  he  left? 

6.  A  boy  earned  x  cents,  found  three  times  as  many,  and 
spent  c  cents.     How  many  cents  had  he  finally  ? 

7.  A  boy  solved  n  examples  and  his  sister  solved  5  more 
than  twice  as  many.  How  many  examples  did  the  sister 
solve  ? 

8.  John  caught  m  trout  and  his  brother  caught  3  less  than 
three  times  as  many.     How  many  did  both  together  catch  ? 

9.  If  William  solves  x  examples,  how  many  examples  must 
John  solve  so  that  both  together  shall  solve  y  examples  ? 


74      THE  LINEAR  EQUATION.  THE  PROBLEM 

10.  Three  men  together  buy  a  field.  B  pays  twice  as  much 
as  A,  C  four  times  as  much  as  B.  If  A  pays  d  dollars,  what 
is  the  cost  of  the  field  ? 

11.  A  horse  cost  y  dollars,  a  harness  x  dollars,  and  a  wagon 
as  much  as  the  combined  cost  of  a  horse  and  two  harnesses. 
What  did  all  three  together  cost  ? 

12.  If  a  line  10  inches  in  length  is  increased  by  n  inches, 
what  is  the  length  of  the  new  line  ? 

13.  How  much  remains  of  a  line  m  inches  long  if  n  inches 
are  cut  from  one  end  andp  inches  are  cut  from  the  other  end  ? 

14.  A  rectangle  is  x  inches  long  and  y  inches  wide,  and  a 
strip  z  inches  wide  is  cut  from  one  end.  What  is  the  area  of 
the  part  cut  off  and  also  of  the  part  remaining  ? 

15.  The  sum  of  two  numbers  is  10  and  the  smaller  number 
is  n.     What  is  the  larger  number  ? 

16.  The  sum  of  three  numbers  is  50;  one  is  x,  and  another 
35.     What  is  the  expression  for  the  third  number  ? 

17.  Name  three  consecutive  numbers,  the  least  of  the  three 
being  n. 

18.  Name  three  consecutive  even  numbers,  the  least  number 
being  n. 

19.  Name  three  consecutive  odd  numbers,  the  least  of  the 
three  being  m. 

20.  Name  five  consecutive  numbers,  the  middle  one  being  x. 

21.  If  x  is  an  even  number,  what  is  the  next  odd  number 
above  x ? 

22.  If  a;  is  an  odd  number,  what  is  the  next  even  number 
above  x ? 

23.  Name  the  three  consecutive  odd  numbers  below  n,  n 
being  an  odd  number. , 

24.  Name  the  three  consecutive  odd  numbers  below  n,  n 
being  an  even  number. 


LITERAL  SYMBOLS  FOR  UNKNOWN  QUANTITIES       75 

25.  What  is  the  sum  of  the  three  consecutive  even  numbers 
below  x  +  1,  x  +  1  being  odd  ? 

26.  If  x  is  the  middle  one  of  three  consecutive  odd  numbers, 
what  is  the  expression  for. their  indicated  product? 

27.  If  a  man  is  x  years  old,  how  old  will  he  be  in  a  years  ? 

28.  If  a  man  is  x  years  old  now,  how  many  years  will  pass 
before  he  is  a  years  old  ? 

29.  If  a  man  is  y  years  old  now,  how  old  was  he  z  years 
ago? 

30.  What  is  the  sum  of  the  ages  of  a  father  and  son  if  the 
son  is  x  years  old  arid  the  father  m  times  as  old  as  the  son  ? 

31.  A  man  is  m  times  as  old  as  his  son  and  n  times  as  old  as 
his  daughter.  If  the  daughter  is  x  years  old,  what  is  the  sum 
of  the  ages  of  all  three  ? 

32.  In  a  period  of  c  year's  a  man  earns  d  dollars.  If  he 
spends  n  dollars  each  one  of  the  c  years,  what  is  the  expression 
for  his  final  saving  ? 

33.  How  many  cents  are  represented  by  x  dimes  ? 

34.  Express  the  condition  that  x  dimes  shall  equal  y  nickels. 

35.  One  dollar  is  lost  from  a  purse  that  had  contained  x 
dollars  and  y  quarters.  Write  the  expression  for  the  amount 
remaining  in  cents. 

36.  A  man  travels  a  miles  an  hour  for  h  hours.  What  is  the 
total  distance  he  travels  ? 

37.  A  boy  rides  x  miles  in  a  train,  then  y  miles  by  boat, 
and,  finally,  by  automobile  twice  as  far  as  he  has  already 
traveled.  Write  the  expression  for  the  total  number  of  miles 
in  his  journey. 

38.  What  is  the  expression  for  the  statement  that  the 
square  of  the  difference  of  two  numbers,  a  and  b,  is  2  less  than 
the  sum  of  the  squares  of  two  other  numbers,  m  and  n  ? 


76      THE  LINEAR  EQUATION.  THE  PROBLEM 

39.  A  room  is  x  yards  long  and  y  yards  wide.  What  is  the 
expression  for  its  area  in  square  yards  ?  in  square  feet  ? 

40.  If  a  man  is  now  x  +  1  years  old,  how  many  years  ago 
was  he  30  years  old?  How  many  years  must  pass  before  he 
will  be  50  years  old  ?  When  was  he  x  —  1  years  old  ?  When 
will  he  be  2  a?  years  old  ? 


PROBLEMS  LEADING  TO  LINEAR  EQUATIONS 

106.  From  the  oral  work  j  ust  covered  the  student  will  see  that 
no  general  directions  can  be  given  for  the  statement^of  problems. 
The  following  suggestions  will  be  of  assistance  in  the  state- 
ment of  all  problems,  and  will  serve  as  a  general  outline  of  the 
method  by  which  we  attack  the  different  types. 

107.  In  stating  a  problem : 

1.  Study  the  problem  to  find  that  number  whose  value  is  re- 
quired. 

2.  Represent  this  unknown  number  or  quantity  by  any  conven- 
ient literal  symbol. 

3.  The  problem  will  state  certain  existing  conditions  or 
relations.  Express  those  conditions  in  terms  of  your  literal 
symbol. 

4.  Some  statement  in  the  problem  will  furnish  a  verbal  equation. 
Translate  this  verbal  equation  into  algebraic  expression  by  means 
of  your  stated  conditions.  The  following  illustrations  will 
show  the  ease  with  which  certain  common  words  and  phrases 
may  be  translated  into  the  common  operations  of  algebra. 

Illustrations : 

1.  The  greater  of  two  numbers  is  3  more  than  the  less,  and 
four  times  the  less  number  exceeds  twice  the  greater  number 
by  8.     Find  the  two  numbers. 


PROBLEMS  LEADING  TO  LINEAR  EQUATIONS    77 

Let  x  =  the  smaller  number. 

Then  x  +  3  =  the  larger  number. 

Now  the  word  "exceeds,"  in  this  case,  may  be  translated  by  the  sign 
"  — ,"  and  the  word  "  by  "  may  be  translated  by  the  sign  "  =." 

Hence,  we  select  the  conditional  sentence  : 

"  four  times  the  less  exceeds  twice  the  greater  by  8," 
and  translate  :4  a;  —  2(x  +  3)  =8. 

Solving  the  equation  4x  —  2(x  +  3)  =  8, 

x  =  7,  the  smaller  number. 

Consequently,  adding  3,  x  +  3  =  10,  the  larger  number. 

The  results  verify  in  the  original  condition,  hence  in  the  equation. 

2.   Find  the  number  which,  multiplied  by  4,  exceeds  40  by  as 
much  as  the  number  itself  is  less  than  40. 

Let  x  =  the  number. 

Then  4  x  =  four  times  the  number. 

Translating  the  conditional  sentence  : 
"  number  .  .  .  multiplied  by  4  exceeds  40  by  as  much  as  40  exceeds  the  number," 
x  X  4-40  =        40      -  x. 

Solving  the  equation,       4  x  —  40  =  40  —  as, 

x  =  16,  the  number. 
Verification : 


4(16)  - 

-40: 

=  40- 

-16. 

64- 

-40 

=  40- 

-16. 

24: 

=  24. 

3.  A  father  is  four  times  as  old  as  his  son,  but  in  24  years 
the  father  will  be  only  twice  as  old  as  the  son.  What  is  the 
present  age  of  each  ? 

Let  x  =  the  number  of  years  in  the  son's  present  age. 

Then  4  x  =  the  number  of  years  in  the  father's  present  age. 

Therefore,      x  +  24  =  the  son's  age  after  24  years. 

4  x  -f  24  =  the  father's  age  after  24  years. 
Now,  "...  in  24  years  the  father  will  be  only  twice  as  old  as  the  son." 
Or,  -24    +  Ax  =  2(«  +  24). 

Solving  the  equation,       24  +  4  x  =  2(x  +  24). 

x  =  12,  the  son's  age  now. 
4  x  =  48,  the  father's  age  now. 
Verification : 

In  24  years  the  father  will  be  (48  +  24)  =  72  years  of  age. 
In  24  years  the  son  will  be       (12  +  24)  =  36  years  of  age. 


78      THE  LINEAR  EQUATION.  THE  PROBLEM 

The  problems  in  the  following  exercise  are  at  first  classi- 
fied in  four  groups  involving  only  the  simplest  of  commonly 
occurring  conditions. 

Exercise  17 
(I)  Problems  involving  One  Number. 

1.  Four  times  a  certain  number  is  36.     Find  the  number. 

Let  x  =  the  required  number. 

From  the  problem  4  x  =  four  times  that  number. 

But  the  problem  states  that 

36  =  four  times  that  number. 
Hence,  from  our  assumed  condition  and  from  the  given  condition,  we 
have  two  expressions,  4  x  and  36,  representing  the  same  quantity. 
Therefore,  4  x  =  36. 

x  =  9,  the  required  number. 
Verification  :  4(9)  =  36. 

36  =  36. 

2.  What  is  that  number  which,  when  decreased  by  5,  gives  a 
remainder  of  19  ? 

3.  Three  times  a  certain  number  is  diminished  by  7  and 
the  remainder  is  11.     What  is  the  number  ? 

4.  William  has  three  times  as  many  books  as  John,  and 
both  together  have  32  volumes.     How  many  has  each  ? 

5.  If  four  times  a  certain  number  is  added  to  five  times  the 
same  number,  the  sum  is  36.     What  is  the  number? 

6.  Four  times  a  certain  number  is  subtracted  from  eleven 
times  the  same  number,  and  the  remainder  is  42.  Find  the 
number. 

7.  I  double  a  certain  number  and  subtract  7  from  the  result, 
and  my  remainder  is  1  more  than  the  original  number. 
What  was  the  number  ? 

8.  If  a  certain  number  is  increased  by  5,  the  sum  is  8  less 
than  twice  the  original  number.     Find  the  number. 


PROBLEMS  LEADING  TO  LINEAR  EQUATIONS  79 

9.  Twelve  times  a  certain  number  is  decreased  by  5,  and 
the  remainder  is  15  more  than  seven  times  the  original  num- 
ber.    What  was  the  number  ? 

10.  Find  that  number  which,  if  doubled,  exceeds  60  by  as 
much  as  the  number  itself  is  less  than  75. 

11.  What  number  is  that  which,  if  doubled  and  subtracted 
from  50,  gives  a  remainder  5  less  than  three  times  the  origi- 
nal number  ? 

(II)   Problems  involving  Two  or  More  Numbers. 

12.  The  sum  of  two  numbers  is  24,  and  the  greater  number 
is  3  more  than  twice  the  smaller  number.     Find  the  numbers. 

Let  x  =  the  smaller  number. 

Then  24  —  x  =  the  larger  number. 

Now  "the  greater  number  is  3  more  than  twice  the  smaller," 


Hence, 

24  -  x  =  3  +  2  x. 

Solving, 

x  =  7,  the  smaller  number. 

Whence, 

24  —  7  =  17,  the  larger  number. 

Verification : 

24  -  7  =  3  +  2(7). 

17  =  3  +  14. 

17  =  17. 

13.  One  number  exceeds  another  number  by  5,  and  their 
sum  is  49.     Find  the  numbers. 

14.  One  number  is  four  times  as  large  as  a  second  number, 
and  their  sum  is  21  more  than  twice  the  smaller  number. 
Find  the  numbers. 

15.  The  sum  of  three  numbers  is  108.  The  second  number 
is  twice  the  first  number,  and  the  third  is  equal  to  twice  the 
sum  of  the  first  and  second.     Find  the  three  numbers. 

16.  Find  the  three  consecutive  numbers  whose  sum  is  54. 

17.  Find  the  three  consecutive  odd  numbers  whose  sum  is  39. 

18.  Find  the  five  consecutive  odd  numbers  whose  sum  shall 
be  equal  to  nine  times  the  smallest  of  the  numbers. 


80      THE  LINEAR  EQUATION.  THE  PROBLEM 

19.  Find  four  consecutive  odd  numbers  such  that  twice  the 
sum  of  the  three  smallest  shall  be  15  more  than  three  times 
the  greatest  one. 

20.  Divide  17  into  two  parts  such  that  the  smaller  part  plus 
four  times  the  larger  part  shall  be  50. 

(Hint  :  Let  x  =  the  smaller  part  •  17  —  x  =  the  larger  part.) 

21.  Divide  64  into  two  parts  such  that  three  times  the 
smaller  part  added  to  twice  the  larger  part  shall  be  158. 

22.  Divide  100  into  two  parts  such  that  twice  the  larger 
part  shall  be  50  more  than  three  times  the  smaller  part. 

23.  Divide  75  into  two  parts  such  that  three  times  the 
larger  part  decreased  by  6  shall  equal  four  times  the  smaller 
part  increased  by  9. 

(Ill)   Problems  involving  the  Element  of  Time. 

24.  A  man  is  twice  as  old  as  his  brother,  but  5  years  ago 
he  was  three  times  as  old.     Find  the  present  age  of  each. 

Let  x  =  the  number  of  years  in  the  brother's  present  age. 

Then  2  x  =  the  number  of  years  in  the  present  age  of  the  man. 

Now  x  —  5  =  the  brother's  age  5  years  ago. 

And        2x  —  5  =  the  man's  age  5  years  ago. 
From  the  statement  in  the  problem : 

2x-5  =  3(x-5). 
Solving,  x  =  10,  the  brother's  age  now. 

2  x  =  20,  the  man's  age  now. 

25.  A  boy  is  5  years  older  than  his  sister,  and  in  4  years 
the  sum  of  their  ages  will  be  29  years.  Find  the  present  age 
of  each. 

26.  A  man  is  twice  as  old  as  his  son,  but  10  years  hence  the 
sum  of  their  ages  will  be  83  years.  What  is  the  present  age 
of  each  ?  i 

27.  Five  years  ago  the  sum  of  the  ages  of  A  and  B  was  50 
years,  but  at  the  present  time  B  is  four  times  as  old  as  A. 
How  old  is  each  now  ? 


PROBLEMS  LEADING  TO  ^LINEAR  EQUATIONS  81 

28.  In  7  years  the  sum  of  the  ages  of  A  and  B  will  be  26 
years  less  than  three  times  A's  present  age.  If  A  is  now  three 
times  as  old  as  B,  find  the  age  of  each  after  7  years. 

29.  In  7  years  a  'boy  will  be  twice  as  old  as  his  brother, 
and  at  the  present  time  the  sum  of  their  ages  is  13  years. 
Find  the  present  age  of  each. 

30.  A  boy  is  three  times  as  old  as  his  sister,  but  in  4 
years  he  will  be  only  twice  as  old.  What  is  the  present  age 
of  each  ? 

31.  A  young  man  is  23  years  of  age  and  his  brother  is  11 
years  old.  How  many  years  ago  was  the  older  brother  three 
times  as  old  as  the  younger  ? 

32.  A  man  50  years  old  has  a  boy  of  9  years.  In  how  many 
years  will  the  father  be  three  times  as  old  as  the  son  ? 

33.  The  sum  of  the  present  ages  of  a  man  and  his  son  is  60 
years,  and  in  2  years  the  man  will  be  three  times  as  old  as 
the  son.  What  will  be  the  age  of  each  when  the  sum  of  their 
ages  is  100  years  ? 

(IV)   Problems  involving  the  Element  of  Value. 

34.  A  man  pays  a  bill  of  $49  with  five-dollar  and  two-dol- 
lar bills,  using  the  same  number  of  each  *kind.  How  many 
bills  of  each  kind  are  used  ? 

Let  x  =  the  number  of  bills  of  each  kind. 

Then  5  x  =  the  value  of  the  fives  in  dollars, 

and  2x  =  the  value  of  the  twos  in  dollars. 

Therefore,  7  x  =  49. 
From  which  x  =  7,  the  number  of  bills  of  each  kind. 

35.  Divide  $100  among  A,  B,  and  C,  so  that  B  shall  receive 
three  times  as  much  as  A,  and  C  $20  more  than  A  and  B 
together. 

36.  A  has  $  16  less  than  B,  and  C  has  as  many  dollars  as  A 
and  B  together.  All  three  have  $  60.  How  many  dollars  has 
each? 

SOM.   EL.    ALG.  —  6 


82      THE  LINEAR  EQUATION.  THE  PROBLEM 

37.  A  number  of  yards  of  cloth  cost  $  3  per  yard,  and  the 
same  number  of  yards  of  silk,  $  7  per  yard.  The  cost  of  both 
pieces  was  $  100.     How  many  yards  were  there  in  each  piece  ? 

38.  $41  was  paid  to  16  men  for  a  day's  work,  a  part  of  the 
men  receiving  $2  per  day  and  the  other  part  $3  per  day. 
How  many  men  worked  at  each  rate  ? 

39.  A  boy  has  $42  in  two-dollar  bills  and  half  dollars,  and 
there  are  three  times  as  many  coins  as  bills.  How  many  has 
he  of  each  kind  ? 

40.  $2.10  was  paid  for  8  dozen  oranges,  part  costing  20  cents 
a  dozen  and  part  costing  30  cents  a  dozen.  How  many  dozen 
were  there  in  each  lot  ? 

41.  A  merchant  bought  50  postage-stamps,  the  lot  being 
made  up  of  the  five-cent  and  the  two-cent  denominations. 
Twice  the  cost  of  the  two-cent  stamps  was  48  cents  more  than 
three  times  the  cost  of  the  five-cent  stamps.  How  many  of 
each  kind  were  bought,  and  what  was  the  total  amount  paid 
for  them  ? 

(V)  Miscellaneous  Problems. 

42.  Find  the  two  numbers  whose  sum  is  70  and  whose  dif- 
ference is  6. 

43.  A  and  B  together  have  $90,  and  A  has  $12  more  than 
B.     How  many  dollars  has  each? 

44.  One  number  exceeds  another  by  3,  and  the  difference 
between  their  squares  is  51.     What  are  the  numbers  ? 

45.  The  difference  between  the  ages  of  a  father  and  son  is 
36  years,  and  the  father  is  three  times  as  old  as  the  son.  Find 
the  age  of  each. 

46.  Divide  70  into  two  parts  such  that  ten  times  the  smaller 
part  shall  equal  eight  times  the  larger  part. 

47.  A  man  divided  $1500  among  four  sons,  each  receiving 
$  50  more  than  the  next  younger.     How  much  did  each  receive  ? 


PROBLEMS   LEADING   TO  LINEAR   EQUATIONS  83 

48.  Divide  31  into  two  parts  such  that  1  less  than  eight 
times  the  smaller  part  shall  equal  five  times  the  greater  part. 

49.  Ten  times  a  certain  number  is  as  much  above  77  as  43 
is  above  five  times  the  number.      What  is  the  number  ? 

50.  The  difference  between  the  squares  of  two  consecutive 
even  numbers  is  52.     Find  the  numbers. 

51.  The  sum  of  two  numbers  is  16  and  the  difference  of  their 
squares  is  32.     What  are  the  numbers  ? 

52.  One  number  is  three  times  another,  and  the  remainder 
when  the  smaller  is  subtracted  from  19  is  the  same  as  the  re- 
mainder when  the  larger  is  subtracted  from  43.  Find  the  two 
numbers. 

53.  A  man  has  three  hours  at  his  disposal  and  walks  out 
into  the  country  at  a  rate  of  4  miles  an  hour.  How  many  miles 
can  he  walk  so  that,  by  returning  on  a  trolley  car  at  the  rate 
of  12  miles  an  hour,  he  will  return  within  just  3  hours  ? 

54.  A  walks  over  a  certain  road  at  a  rate  of  3  miles  an  hour. 
Two  hours  after  he  leaves,  B  starts  after  him  at  a  rate  of  4 
miles  an  hour.  How  many  miles  will  A  have  gone  when  B 
overtakes  him  ? 

55.  A  and  B  are  60  miles  apart  and  start  at  the  same  time 
to  travel  towards  each  other.  A  travels  4  miles  an  hour  and 
B  5  miles  an  hour.  In  how  many  hours  will  they  meet  and 
how  far  will  each  have  traveled  ? 

56.  A  man  walks  5  miles  on  a  journey,  rides  a  certain  dis- 
tance, and  then  takes  an  automobile  for  a  distance  four  times 
as  great  as  he  has  already  traveled.  In  all  he  travels  75  miles. 
How  far  does  he  go  in  the  automobile  ? 

57.  How  can  you  pay  a  bill  of  $5.95  with  the  same  number 
of  coins  of  each  kind,  using  only  dimes  and  quarters? 

58.  The  sum  of  the  ages  of  a  father  and  son  is  96  years ; 
but  if  the  son's  age  is  trebled,  it  will  be  8  years  greater  than 
the  father's  age.     How  old  is  each  ? 


CHAPTER  VII 

SUBSTITUTION 

108.   Substitution  is  the  process  of  replacing  literal  factors 
in  algebraic  terms  by  numerical  or  by  other  literal  values. 
Illustrations : 
1.   If  a  =  5  and  6  =  7: 


(«  +  &)  = 

(5  +  7) 
12.     Result. 

2. 

If  a 

=  4  and  b=  -2: 

(2a-a26)  = 

.  [2-4 -42  (-2)] 
8 -16  (-2) 
8  +  32 
40.     Result. 

3. 

If  a; 

=  4  m,  y  —  3  m,  and 

Z=  —  5m: 

(jc  +  y  -  z)  = 

(4  m  +  3  m  +  5  m) 
12  m.     Result. 

4.   With  the  same  values  for  x,  y,  and  %  as  in  (3)  : 

(aj  +  2/2)(2a;-^-2«2)  =  [4m+(3m)2][2(4m)-3m-2(-5m)2] 
=  (4m  +  9w2)[8ra-3m-2(25m2)] 
=  (4  m  +  9  m2')  (5  m  -  50  m2) 
=  (20  m2  -  155  m8  -  450  wi4).     Result. 

From  these  illustrations  we  state  the  general  method  for 
substitution : 

109.  Replace  the  literal  factors  of  the  terms  of  the  given  ex- 
pression by  their  respective  given  values.  Perform  all  indicated 
operations  and  simplify  the  result. 

84 


SUBSTITUTION  OF  NUMERICAL  VALUES  85 

I.    SUBSTITUTION  OF  NUMERICAL  VALUES 

Exercise  18 

Find  the  numerical  values  of  the  following  when  a  =  4,  6  =  3, 
c  =  4,  and  d  =  1. 

1.  a+b  +  c.  7.  ab  —  3  be  +  5  ad. 

2.  a-26  +  5c.  8.  2ab-3cd  +  2bd. 

3.  46-3c  +  2a\  9.  5ad,-36a,  +  8cd'. 

4.  7a—  a"  +  9c  —  6.  10.  ab—(bc—cd). 

5.  10a  +  6-(3c-d).  11.  8a6c-5  6ca'  +  2aca\ 


6.   46  —  [3  c  —  (2a  +  a")].        12.   acd*  —  (abc  —  bed  —  acd). 
13.   acd-2  6cd-3(a6-c(f). 


14.  2a6-  (3cd  +  2a  —  a6  +  6c  —  bd). 

15.  6c-[-6d-(a6-cd")  +  (a6  — 6d*)]. 

16.  Simplify  a(a+  c)  —c(c  —  a)  when  a  =  4  and  c  =  2. 

17.  Simplify  (a  +  6)2  -  (a  +  6) (a  -  b)  -  (a  -  6)2  when  a  =  5 
and  6  =  4. 

18.  Simplify  10  (a  -f  l)2  -  (3  a  -  2)2  when  a  =  -2. 

19.  Simplify  4 (2  a;  -f-  y)2  —  S(x  +y) (x  —  y)  when  x  =  —  2  and 
2/  =  0. 

20.  Simplify  3(ra  -2x)2-  2(ra  +  2  »)(m  +  3  x)  -f  4  x  when 
m  =  —  5  and  a;  =  2. 

21.  Simplify  d  —  2(c  +  d)(c  -  2  d)  -  3(e  -  d)  when  c  =2  and 
a"  =-2. 

22.  Simplify  a(a  +  6)  -  b(a  -  6)2  -  (a  +  &)3  when    a  =  -  3 
and  6  =  —  2. 

23.  Simplify  (3  a  +  m)(a-  5  m)  +[9  a2-  |ra-a(2m-9a)  j] 
when  a  =  0  and  m  =  10. 


86  SUBSTITUTION 

24.  Simplify  7  a(a  + 1)  -  (2  +  a)  (3  a  -  5)  -  3  a(a  - 1)  when 
a  =  0. 

25.  Simplify  (3  m-  2)2  +(w  +  n-l)»-(2w-  1)  (m  + 1) 
when  m  =  n=  —  3. 

II.    SUBSTITUTION  OF  LITERAL  VALUES 

When  literal  values  are  substituted,  the  results  will  be  in 
terms  of  those  literal  values. 

Exercise  19 

Simplify : 

1.  (a  —  x)2-\-(a  —  2x)2  when  a  =  2x. 

2.  (a  —  3  x  —  x2)  when  x  =  2a. 

3.  (a  +  x)(a  —  2x)2—  (a  —  x)2  when  a=  —  x. 

4.  (a?  +  n)(x  +  ft  —  1)  when  x  =  m  and  n  =  —  m. 

5.  (2  a  —  3 c  -\-  x)(x  —  2 a)  —  2  acx  when  or  =  x  and  c  =  0. 

6.  5  ac  —  (a  +  c)(a  —  4  c)  -f-  (a  —  c)2  when  c  =  0. 

7.  2(ra  —  a)  —  2 (m  +  3 x)(m  —  2  x)  —  3  mx  when  ra  =  a  and 
x  =  —  2  a. 

8.  (2a>+l)*-3a?-4a>(a!-5)  whena=  -2a. 

9.  (3x-l)2-(3x+l)(2x-5)  whenx  =  ab. 

10.  (3a2-2a  +  l)-(3a;-2)2when  »=  —  mnl 

11.  (m  +  a?)(m  —  #)  —  (m  —  a;)2  +  2 m  when  m=  —x. 

12.  (a-26  +  8)2-x(a;  +  2a)-(a-2&)3-l  when  a  =  2& 
and  x  =  0. 

13.  (6  ar>  - 11  o?y  -  10  y2)  -  6(a^  —  2xy)  +  11 2/2  when  aj  =  m 
and  y——m. 

14.  (a?  —  a)(aj  +  a  —  y)  —  (x  —  y)  (x  -\»  a)  a  —  (x  —  a)2  when 
x=  —  m  and  a  =  0. 

15.  c(c  +  c7  +  j)  —  (c  —  d-f-l)(c+d— 1)  —  cd— 1  whenc=d=0. 


THE   USE   OF   FORMULAS  87 

HI.    THE   USE  OF  FORMULAS 

110.  A  formula  is  an  algebraic  expression  for  a  general 
principle.  For  example  :  If  the  altitude 
of  a  triangle  is  represented  by  a,  the  base 
by  b,  and  the  area  by  S,  we  have  the 
general  expression  for  its  area  in  the 
formula 

o      ab 

Given  the  values  of  a  and  b  for  a  par- 
ticular triangle,  we  obtain  its  area  by  substituting  those  values 
in  this  general  formula  and  simplifying. 

A  few  common  formulas  will  illustrate  the  value  of  this  brief 
method  of  expression. 

(1)  The  Formula  for  Simple  Interest. 

By  arithmetic :  The  interest  (I)  on  a  given  principal  at  a 
given  rate  for  a  given  time  equals  principal  x  rate  x  time.  Ex- 
pressed as  a  formula: 

If  p  =  the  principal  expressed  in  dollars, 
r  =  the  rate  of  interest  expressed  decimally, 
t  =  the  time  expressed  in  years, 
Then        /  =  prt  is  the  general  formula  for  simple  interest. 

(2)  The  Formula  for  Compound  Interest.  (Interest  com- 
pounded annually.) 

If  p  =  the  principal  expressed  in  dollars, 

r  =  the  rate  of  interest  expressed  decimally, 
n  =  the  number  of  years  in  the  interest  period, 
Then        /=p(l  +r)n  —  p  is  the  general  formula  for  interest  com- 
pounded annually. 

(3)  The   Formula  for  the    Transformation    of    Temperatures. 

Both  the  standard  thermometers,  the  Fahrenheit  and  the  Centi- 
grade, are  in  everyday  use  in  physical  investigations,  and  the 
formula  given  below  is  used  to  change  Fahrenheit  readings  to 
the  Centigrade  scale. 


88  SUBSTITUTION 

If  F  =  the  given  reading  from  a  Fahrenheit  scale, 

C  =  the  required  equivalent  reading  on  the  Centigrade  scale, 
Then  C  sa  $  (F  —  32)  is  the  formula  for  temperature  transformation. 

Exercise  20 

1.  What  is  the  area  of  a  triangle  having  a  base  of  14  inches 
and  an  altitude  of  9  inches  ? 

2.  Find  the  simple  interest  on  $  1500  for  12  years  at  5  per 
cent. 

3.  What  is  the  interest  on  $6200  for  3  years  7  months 
10  days  at  4.5  per  cent  ? 

4.  Find  the  interest,  compounded  annually,  on  $  1200  for 
4  years  at  5  per  cent. 

5.  On  a  Fahrenheit  thermometer  a  reading  of  50°  is  taken. 
What  is  the  equivalent  reading  on  a  Centigrade  thermometer? 

6.  Find  the  circumference  (C)  of  a  circle,  the  radius  (R) 
being  5  feet,  and  the  constant  (it)  in  the  formula  0=2  ttR 
being  3.1416  +  . 

7.  Find  the  area  (S)  of  a  circle  whose  radius  (R)  is  4  feet, 
the  constant  (71-)  in  the  formula  S  =  irR2  being  3.1416  +. 

8.  With  the  formula  of  Example  7,  find  the  area  of  a  cir- 
cular pond  whose  diameter  (D)  equals  100  feet.     (D  —  2  R.) 

9.  If  the  diameter  (D)  of  a  sphere  is  2  feet,  what  is  the 

volume  (V)  of  the  sphere  from  the  formula,  V— ?     (tt  = 

3.1416  +.)  6 

10.  Find  the  last  term  (Z)  in  a  series  of  numerical  terms  of 
which  the  first  term  (a)  is  3,  the  number  of  terms  (n)  is  8, 
and  the  difference  between  the  terms  (d)  is  2,  the  formula  be- 
ing l  =  a+  (n  —  l)d. 

11.  If  a  =  3,  r  =  2,  and  n  m  5,  what  is  the  value  of  I  in  the 
expression  I  =  arn_1  ? 

12.  If  r  =  5,  s  =  4,  and  n  =  6,  find  the  value  of  a  in  the  ex- 
pression a  =  (r  —  l)s  -f-  rn_1. 


CHAPTER   VIII 


SPECIAL  CASES  IN  MULTIPLICATION  AND  DIVISION 

MULTIPLICATION 

The  product  of  simple  forms  of  binomials  may  often  be 
written  without  actual  multiplication,  the  result  being  obtained 
by  observing  certain  laws  seen  to  exist  in  the  process  of  actual 
multiplication.     Three  common  cases  are : 


The  square  of  the 
sum  of  two  quan- 
tities, 
a  +      b 
a  +      b 
a2  +    ab 
+     ab  +  b2 


II 

The  square  of  the 
difference  of  two 

quantities, 
a  —      b 
a  —      b 
a2-     ab 

-     ab  +  b2 


III 

The  product  of  the 

sum  and  difference 

of  two  quantities. 

a  +    b 

a  —   b 

a2  +  ab 
-ab-b2 


a2 


F 


a2+  2  ab  +  b2  a2  -  2  ab  +  b2 

Therefore,  we  may  state,  from  (I),  (II),  and  (III), 
respectively : 

111.  The  square  of  the  sum  of  two  quantities  equals  the  square 
of  the  first,  jplus  twice  the  product  of  the  first  by  the  second,  plus 
the  square  of  the  second. 

112.  The  square  of  the  difference  of  two  quantities  equals 
the  square  of  the  first,  minus  twice  the  product  of  the  first  by  the 
second,  plus  the  square  of  the  second. 

113.  The  product  of  the  sum  and  difference  of  two  quantities 
equals  the  difference  of  their  squares. 

Ability  to  apply  these  principles  rapidly  is  essential  in  all 
later  practice. 

89 


90       SPECIAL  CASES  IN  MULTIPLICATION  AND  DIVISION 

Oral  Drill 

Give  orally  the  products  of  : 

1.  (a  +  m)2.  4.    (a^+4)2.  7.  (3a  +  2)2. 

2.  (x  +  z)2.  5.    (a  +  3m)2.  8.  (4a  +  5)2. 

3.  (a;  +  3)2.  6.    (c+5d)2.  9.  (5c+7)2. 

10.  (c-z)2.  13.    (c-8)2.  16.  (cd-4)2. 

11.  (&-4)2.  14.    (3a-5)2.  17.  (a262-6)2. 

12.  (m-5)2.  15.    (7a-3)2.  18.  (cd-Scx)2. 


19.  (a  +  a;)(a-aj).  27.  (3  mn  +  5  ma)2. 

20.  (m  +  y)(m-y).  28.  (4c2d2-5)2. 

21.  (a +  4) (a- 4).  29.  (2m3  +  9)(2ra3-9). 

22.  (2a  +  l)(2a-l).  30.  (5  xyz  +  7  y2z2)2. 

23.  (3a  +  5)(3a-5).  31.  (3a4-13)(3a;4  +  13). 

24.  (7a  +  10)(7a  — 10).  32.  {am2 +  xyz)(am2 -xyz). 

25.  (5a-8  2/)(5a;  +  8  2/).  33.  (3c5  +  ll)2. 

26.  (3  a2 +  5) (3  a2 -5).  34.  (7  m6 -9  a7)2. 

IV.    THE  DIFFERENCE  OF  TWO  SQUARES  OBTAINED  FROM  TRINOMIALS 

114.  Many  products  of  two  trinomials  may  be  so  written 
as  to  come  under  the  binomial  case  of  Art.  113.  In  such 
multiplications  we  group  two  of  the  three  terms  in  each  quantity 
so  as  to  produce  the  same  binomial  expressions  in  each,  the  pa- 
renthesis being  treated  as  a  single  term.  Three  different  cases 
may  occur : 

(1)  (2)  (3) 

(a +  6  +  c)(a  +  b-c)       (a  +  b  +  c)(a-  b  +  c)        (a  +  6-c)(a  -b  +  c) 

The  terms  inclosed  in  parentheses  must  form  the  same  binomial 
in  each  expression. 


THE  DIFFERENCE   OF  TWO   SQUARES  91 

From  1.      (a  +  6  +  c)(a  +  b  -  c)  =  [(a  +  b)  +  e][(a  +  6)  -  c] 

=s  (a  +  &)2  -  c2 

=  a2  +  2  a&  +  62  -  c2.     Result. 

From  2.      (a  +  b  +  c) (a  -  b  +  c)  =  [(a  +  c)  +  6] [(a  +  c)  -  6] 

=  (a  +  c)2  -  62 
=  a2  +  2  ac  +  c2  -  62.     Result. 

From  3.  In  this  case  only  one  term  has  the  same  sign  in  each  ex- 
pression, the  a-term.  Hence  the  last  two  terms  of  each  expression  ara 
inclosed  in  parentheses,  the  parenthesis  in  one  case  being  preceded  by 
the  minus  sign. 

(a  +  b  -  c)(a  -  b  +  c)  =  [a  +  (6  -  c)][a  -  (6  -  c)] 
=  a2  -  (6  -  c)2 
=  a2  -  (62  -  2  6c  +  c2) 
=  a2  -  62  +  2  be  -  c2.     Result. 

Exercise  21 
Write  by  inspection  the  following  products  : 

1.  [(a  +  0)  +  4][(a+«)— 4].      8.    (ra  +  a  +  2)(ra  +  a;  —  2) 

2.  [(ra  +  a;)  +  2][(m  +  a)—  2].     9.   (m  —  w  +  c)(ra  —  n  —  c). 

3.  [(c-2)  +  m][(c-2)-m].     10.    (c2  +  c  +  l)(c2  +  c-l). 

4.  [(a2  +  l)  +  a][(a2  +  l)-a].   11.    (c-d  +  2)(c-d-2). 

5.  [a -r-(d  +  a>)] [a —  (d  +  a;)].      12.    (m  +  w  — 4)(m  — n  +  4), 

6.  [d+(y-«)][<i-(y-«)3.        13.    (^-2-^(^-2  +  ^), 

7.  (a+a.  +  y)(a  +  a;-^).  14>    (x*jtX-2){xi-x  +  2). 

15.  (aj4  +  ^  +  l)(»4-^+l). 

16.  (m4-2m2  +  l)(m4  +  2m2  +  l). 

17.  (x4-x2-6)(xA  +  x2-6). 

18.  [(c  +  d)  +  (m  +  l)][(c  +  d)-(m  +  l)]. 

19.  [0  +  x)  -  (y  -  z)]  [(«  +  <*)  +  &-  *)J 

20.  (t^-^-c-lX^  +  ^  +  c-l). 


92       SPECIAL   CASES   IN  MULTIPLICATION  AND   DIVISION 

V.    THE  PRODUCT  OF  TWO  BINOMIALS  HAVING  A  COMMON 
LITERAL  TERM 

By  actual  multiplication  :        a  +    7 

a  +    5 


a2  +    7  a 

+    5a +  35 
a2  +  12  a  +  35 
Hence,  in  the  product : 

a?  =  a  x  a,  the  product  of  the  given  first  terms. 

+  12a=(+7  +  5)a     the  product  of  the  common  literal  term  by  the 

sum  of  given  last  terms. 
+  35  =  (+  6)  (+  7)         the  product  of  the  given  last  terms. 

In  like  manner  :  (1 )  (x  -  4)  (x  -  9)  =  z2  - 13  x  +  36. 

x*  =  x-x.     -13z=(-4-9)z.     +  36=(-4)(-9). 

(2)  (e  +  9)(z-3)=z2  +  6z-27. 
x2  =  x-x.      +6x=(+9-3)x.     -27=(+9)(-3). 

(3)  (w-15)(m+7)=m2-8m-105. 
ma  =  OT-m.    -8w=(-15  +  7)to.    -105=(- 15)(+7) 

In  general :  (x  +  a)  (x  +  b)  =  «2+  (a  +  b)x  +  ab. 

Therefore,  in  the  product  of  two  binomials  having  a  com- 
mon literal  term: 

115.  The  first  term  is  the  product  of  the  given  first  terms. 
The  second  term  is  the  product  of  the  common  literal  term  by  the 
algebraic  sum  of  the  given  second  terms.  The  third  term  is  the 
product  of  the  given  second  terms. 

Oral  Drill 

Give  orally  the  following  products : 

1.  (a>  +  3)(a>  +  ±).  6.    (c  +  l2)(c  +  l). 

2.  («  +  4)(o;  +  5).  7.   (tt  +  ll)(n  +  12). 

3.  (a>  +  5)(»  +  7).  8.    (m  +  3)(m  +  20). 

4.  (m  +  6)(m  +  4).  9.   (d  +  12)(d  +  16). 

5.  (y +  9)0/ +  4).  10.    (c  +  15)(c  +  16). 


THE   PRODUCT  OF  ANY  TWO  BINOMIALS  93 

11.  (6-4)(6-7).  15.  (xz-9)(xz-l). 

12.  (n-ll)(n-10).  16.  (c2  -  3) (c2  -  4). 

13.  (ax-3)(ax  —  5).  17.  (x2  -  xy)(x>  -  2  xy). 

14.  (cd—  T)(cd  —  3).  18.  (mn-  3n)(mw-  7  w). 

19.  («-3)(sc  +  5).  24.  (a-13)(a+10). 

20.  (c-3)(c  +  8).  25.  (a2  +  5)(a2-3). 

21.  (x-9)(x  +  7).  26.  (m34-8)(m3-3). 

22.  (a?4-9)(a?-10).  27.  (a&  -  9)(ao  + 12). 

23.  (m  +  H)(m-12).  28.  (cx2-7)(cx2  +  S). 


VI.  THE  PRODUCT  OF  ANY  TWO  BINOMIALS 

By  actual  multiplication 


2a  +5 

3a  +4 

6  a2  +  15  a 

+    8  a -f  20 

6  a2  +  23  a  +  20 

The  two  multiplications  resulting  in  the  terms  + 15  a  and 

+  8  a  are  cross  products.     It  will  greatly  assist  the  beginner 

to  imagine  that  the  terms  entering  the  cross  product  have  this 

connection : 

I r=i 1 

(2a+5)(3a+4) 

And,  by  inspection,  the  middle  term  results  from 

(+2a)(+4)+  (+3a)(+5)  =  +  8  a  +  15  a  =  +  23  a. 
Considering  other  possible  cases  : 

(1)  In  (4  a  —  7)  (3  a  —  5)  the  middle  term  in  the  product  will  be 

(+4a)(-5)  +  (-f3a)(-7)  =  _  20  a  -21  a  =  -  41  a. 

(2)  In  (2  a  +  5)  (3  a  —  4)  the  middle  term  in  the  product  will  be 

(4-  2  a)(-  4)  +  (+  3  «)(+  5)  =  -  8  a  +  15  a  =  +  7  a. 

(3)  In  (3  a  —  2)  (9  a  +  4)  the  middle  term  in  the  product  will  be 

(+3a)(-4)  +  (+9a)(-2)  =  +  12  a  -  18  a  =  -  6  a. 


94       SPECIAL  CASES  IN  MULTIPLICATION  AND  DIVISION 

In  general :   (ax  +  b)  (ex  +  d)  =  acx2  +  (acd  +  bc)x  +  bd. 
Therefore,  in  the  product  of  any  two  binomials : 

116.  The  first  term  is  the  product  of  the  given  first  terms. 
TJie  second  term  is  the  algebraic  sum  of  the  cross  products  of  the 
given  terms.  The  third  term  is  the  product  of  the  given  second 
terms. 

Exercise  22 

Write  by  inspection  the  following  products : 

1.  (3»  +  2)(aJ  +  l).  13.  (7aj-2)(3o?-7). 

2.  (2  a  -fl)  (3  a  +  2).  14.  (10  c- 11)  (3  c- 7). 

3.  (3m+2)(2m  +  3).  15.  (7  n  -9) (8  n-  5). 

4.  (4a?  +  JL)(a  +  3).  16.  (5mw  — 1)(3  mn-4). 

5.  (5c  +  4)(2c  +  3).  17.  (4a  -f  7m)(3a  -  2  m) 

6.  (66  +  5)(26  +  3).  18.  (7 c  -  Sd)(Sc  +  3d). 

7.  (5z  +  7)(3z  +  4).  19.  (2  ac  -  3  a>)(ac  +  11  x) 

8.  (7m  +  2)(2m  +  9).  20.  (14m  —  5 no;) (2m  +  nx). 

9.  (3a-7)(2a-5).  21.  (11  x  +  9y)(9x  -  2  y). 

10.  (3m-l)(2m-3).  22.    (3  c  -  13 mn) (4 c  +  5 mn). 

11.  (5c-2)(3c-l).  23.    (mna-ll)(2mna;  +  3). 

12.  (6y-l)(2y-3).  24.    (13 ac  -  5 x)(3ac  +2 x). 


VII.    THE  SQUARE  OF  ANY  POLYNOMIAL 
By  actual  multiplication : 

(a  +  b  +  c)2  =  a2  +  b2  +  c2  +  2  ab  +  2  ac  +  2  6c. 

(a  +  &  _  c  _  <Z)2  _a2 +  &2  +  c2  +  ^2  +  2a&  _ 2ac  -  2aa*  -  26c  -  26a" +2ca\ 

It  will  be  seen  that,  in  each  product, 

(1)  The  square  terms  are  all  positive  in  sign. 

(2)  The  other  terms  are  positive  or  negative  according  to 
the  signs  of  their  factors. 


THE  DIFFERENCE  OF  TWO   SQUARES  95 

(3)  The  coefficient  of  each  product  of  dissimilar  terms  is  2. 

Applying  the  principle  to  a  representative  example  we  have : 

(a-2&  +  3c)2  =  (a)2  +  (-2  6)2+(3c)2  +  2(a)(-2&)  +  2(a)(+3c)  + 
2(_2&)(+3c)  =  a2  +  462  +  9c2-4a&  +  6  ac-  12  be.    Result. 

In  general : 

117.  The  square  of  any  polynomial  is  the  sum  of  the  squares 
of  the  several  terms  together  with  twice  the  product  of  each  term 
by  each  of  the  terms  that  follow  it. 

Exercise  23 

Write  by  inspection  the  following  products  : 

1.  (a  -\-  m  +  rif.  7.    (a  +  c  +  m  +  x)2. 

2.  (c  +  d  +  x)2.  8.    (m-2w  +  3x-  2)2. 

3.  (a  +  c-m)2.  9.    (3a-26-c-l)2. 

4.  (a +  26  + 3c)2.  10.   ^(l-c-c2-^)2. 

5.  (2m-3rc  +  4)2.  11.   *(m3  -  m2  +  m  -  l)2. 

6.  (3m-4n-5)2.  12.   *(d3  -  3d2  +  4d  -  2)2. 

DIVISION 

In  certain  cases  where  both  dividends  and  divisors  are  bi- 
nomials we  are  able  to  write  the  quotients  without  actual  divi- 
sion. Such  divisions  are  limited  to  the  cases  where  the  terms 
of  the  dividend  are  like  powers ;  that  is,  both  terms  must  be 
squares,  both  cubes,  both  fourth  powers,  etc.  The  powers  of 
the  binomial  divisors  are  also  like. 

I.    THE  DIFFERENCE  OF  TWO  SQUARES 
By  Art.  113,  (a  +  b)(a  -  6)  =  a2  -  62. 

Therefore,  by  division  : 

^^!  =  a_6,         and         «^?  =  a  +  6. 

a  +  b  a  —  b 

*  After  squaring,  the  terms  should  be  collected. 


96       SPECIAL  CASES  IN  MULTIPLICATION  AND  DIVISION 
Hence  the  general  principle  may  be  stated  as  follows : 

118.  The  difference  of  the  squares  of  two  quantities  may  be 
divided  by  either  the  sum  or  the  difference  of  the  quantities.  If 
the  divisor  is  the  sum  of  the  quantities,  the  quotient  will  be  the 
difference  of  the  quantities  ;  and  if  the  divisor  is  the  difference 
of  the  quantities,  the  quotient  will  be  the  sum. 

-  Oral  Drill 
Give  orally  the  quotients  of : 

1.  (a2-ra2)-=-(a-m).  8.    (16-9m*)-s-(4-3m). 

2.  (m* -«■)-*-(«* -a).  9.    (25 x 2 - 49 y*)  +  (5 x-  1  y). 

3.  (a2 -4)+ (a  +  2).  10.    (49  c2cZ2  -  9) -- (7  cd  + 3). 

4.  (aj»_9)-*-(a>  +  3).  "■■    (100 -8 la4) -(10 -9 a2). 

5.  (c2-36)^-(c-6).  12.    (rfy2-  121)  +  (xy  + 11). 

6.  (4m2-l)-f-(2m-l).  13.    (81mW-169c4)-s-(9mw-13c*). 

7.  (9d2-25)-j-(3d+5).  14.   (196c6-81d4)-(14c3  +  9d2). 

II.    THE  DIFFERENCE  OF  TWO  CUBES 

By  actual  division  :       qB  ~  &B  =  a2  +  ab  +  &2. 
a—b 

From  the  form  of  the  quotient  and  its  relation  to  the  divisor 

we  state: 

119.  The  difference  of  the  cubes  of  two  quantities  may  be 
divided  by  the  difference  of  the  quantities.  The  quotient  is  the 
square  of  the  first  quantity,  plus  the  product  of  the  two  quantities, 
plus  the  square  of  the  second  quantity. 

Oral  Drill 

Give  orally  the  quotients  of: 

1.  (a3-c3)-5-(a-c).  3.    (or*  -  z3)  -*-  (x  -  *). 

2.  (m8-a?)-*-(m-a;).  4.    (x*-l)-*-(x-l). 


THE  SUM  OF  TWO  CUBES  97 

5.  (c8_8)  +  (c-2).  9.  (c3d3 -  125)  +  (cd- 5). 

6.  (^_27)_^(d_3).  io.  (8  ar>-  27  m3)  +  (2  x  -3  m). 

7.  (64 -a?) -s- (4  -x).  11.  (512c3d3-729)-=-(8cd-9). 

8.  (ary-l)-r-(a;?/-l).  12.  (216 afy8 - 1000 z3)^ (6ay -10 2). 

m.    THE  SUM  OF  TWO  CUBES 

By  actual  division  :     q3  +  b*  =  a?-ab  +  62. 
a-tb 

From  the  form  of  the  quotient  and  its  relation  to  the  divisor, 
we  state: 

120.  The  sum  of  the  cubes  of  two  quantities  may  be  divided  by 
the  sum  of  the  quantities.  The  quotient  is  the  square  of  the  first 
quantity,  minus  the  product  of  the  two  quantities,  plus  the  square 
of  the  second  quantity. 

Oral  Drill 

Give  orally  the  quotients  of : 

1.  (a*+c*)+<a+c).  7.   (125  +  cP)  +  (5  +  d). 

2.  (m*  +  a?)-*-(m  +  x).  8.    (27  +  d?)  -+-  (3  +  d). 

3.  (c3  +  ^)-(c  +  2/).  9.    (mY  +  l)  +  (my  +  l). 

4.  (a3  +  l)-f-(a  +  l).  10.    (m3x3  +  125)^-(maj  +  5). 

5.  (aj»  +  8)-*-(»  +  2).  11.    (8c»  +  27a?)-*-(2c  +  3*).  ■ 

6.  (n3  +  27)-r-(7i.  +  3).  12.    (125  63  +  64)  -s-(5  b  +  4). 

13.    (729 m3n6  + 1000 a;9)-  (9 mn2  + 10 a3). 

IV.    THE  SUM  OR  DIFFERENCE  OF  AWT  TWO  LIKE  POWERS 

(a)  The  Difference 

Even  Powers  Odd  Powers 

a2  -  62  as  _  &8       , 

a4  —  64        may  each  be  divided  by       a5  —  66        may  each  be  divided 
a6  -  66     (a  +  6)  or  (a  -  6).  a7  -  67    by  (a  -  6)  only. 

a8  -  ft8  a9  _  59 

etc.  etc. 

SOM.    EL.    ALG. 7 


98       SPECIAL  CASES  IN   MULTIPLICATION  AND   DIVISION 


(b)  The  Sum 

Even  Powers 

Odd  Powers 

a2  +  &2 

a8  +  68 

a* +  6* 

are     not     divisible     by 

a5  +  65 

may  each   be  divided 

a6  +  66 

either  (a  +  b)  or  (a  —  6). 

a7  +  67 

by  (a  +  6)  only. 

a8  +  68 

a9+69 

etc. 

etc. 

A  general  proof  for  these  cases  is  considered  in  later  practice. 
Illustrations : 
(1)  a4~&4  =  o8  +  a25  +  a&2  +  68-         (2)  q*  ~  ~  =  a8  -  a26  +  a&2  -  68. 

(3)  «L±A7  =  a6  -  a56  +  a462  -  a363  4-  a2&4  -  a&6  +  6«. 
a  +  b 

From  the  form  of  the  quotients  and  their  relation  to  their 
divisors  we  state : 

121.  The  number  of  terms  in  the  quotient  is  the  same  as  the 
exponent  of  the  powers  in  the  dividend. 

The  exponent  of  "a"  in  the  first  term  of  the  quotient  is  the 
difference  between  the  given  exponents  of  "  a  "  in  the  dividend  and 
divisor,  and  this  exponent  decreases  by  1  in  each  successive  term. 

The  exponent  of  "  b "  is  1  in  the  second  term  of  the  quotient, 
and  this  exponent  increases  by  1  in  each  successive  term  until 
it  is  equal  to  the  difference  of  the  given  exponents  of  "  b  "  in  the 
dividend  and  divisor. 

If  the  sign  of  "b"  in  the  divisor  is  -{-,  the  signs  of  the  quotient 
are  alternately  +  and  — ,  and  if  the  sign  of'b"  in  the  divisor  is 
— ,  the  signs  of  the  quotient  are  all  +• 

Exercise  24 

Write  by  inspection  the  following  indicated  quotients : 
%    aA-b\  f  gt+y5  g    c5  +  1  ?    m^ 


a  —  b  a  +  y  c-fl  m2  —  n 

i5-b5  4    c4-d*  6    xY-z\       g    fl*  +  32 

a—b  c  +  d  xy  —  z  x+   2 


CHAPTER  IX 
FACTORING 

122.  An  algebraic  expression  is  rational  with  respect  to  any 
letter  if,  when  simplified,  the  expression  contains  no  indicated 
root  of  that  letter.     Thus : 

x2  —  xy  +  y/y  is  rational  with  respect  to  x,  but  irrational  with  respect  to  y. 
The  symbol  -y/  is  used  to  indicate  a  required  square  root. 

123.  An  algebraic  expression  is  integral  with  respect  to  any 
letter  if  that  letter  does  not  occur  in  any  denominator.     Thus : 

x  +  —  +  m  is  integral  with  respect  to  m,  but  fractional  with  respect  to  x. 

x 

124.  An  algebraic  expression  is  rational  and  integral  if  it  is 
rational  and  integral  with  respect  to  all  the  letters  occurring 
in  it.     Thus : 

4  m2  —  5  mn  +  se4  is  both  integral  and  rational. 

125.  The  factors  of  a  rational  and  integral  algebraic  ex- 
pression are  the  rational  and  integral  algebraic  expressions 
that,  multiplied  together,  produce  it.     Thus : 

atbZm  =  axaxbxbxbxm, 
and  a,  a,  6,  ft,  b,  and  m  are  the  factors  of  a%zm. 

One  of  two  equal  factors  of  a  number  is  a  square  root  of  the 
number  ;  one  of  three  equal  factors,  a  cube  root ;  of  four  equal 
factors,  a  fourth  root ;    etc.    Thus : 

x  is  the  square  root  of  x2  ;  y  is  the  cube  root  of  ys  ;  etc. 


100  FACTORING 

126.  An  algebraic  expression  is  a  prime  expression  when  it 
has  no  factors  excepting  itself  and  unity.  In  the  process  of 
factoring  we  seek  prime  factors,  and  we  extend  our  work  until 
the  resulting  expressions  have  no  rational  factors. 

Unless  otherwise  stated,  the  use  of  the  term  "algebraic  expression" 
in  elementary  algebra  is  understood  to  refer  to  rational  and  integral  ex- 
pressions only. 

127.  An  expression  is  factored  when  written  in  the  form  of  a 
product. 

128.  Since  factorable  expressions  result  from  some  com- 
pleted multiplication,  we  may  repeatedly  refer  to  multiplication 
as  the  origin  of  particular  type  forms  for  which  we  have  definite 
methods  of  factoring.     The  forms  are  classified  by 

1.  The  number  of  terms  in  the  expression. 

2.  The  powers  and  coefficients  of  the  terms. 

3.  The  signs  of  the  terms. 


WHEW  EACH  TERM  OF  AN  EXPRESSION  HAS  THE  SAME  MONOMIAL  FACTOR 

Type  Form  •  •  •  ax  +  ay  -\-az. 

The  Origin :  If  a  polynomial  (x  +  y  +  z)  is  multiplied  by  a 
monomial  (a),  we  obtain  (Art.  55) : 

a(x  +  y  +  z)  =zaz  +  ay  +  az. 
Therefore,  ax  +  ay  +  o,z  =  a(x  +  y  +  z). 

That  is,  the  factors  of  ax  +  ay  +  az  are  "  a  "  and  "  x  +  y  +  z." 

In  general,  to  factor  an  expression  in  which  each  term  has 
the  same  monomial  factor : 

129.  By  inspection  find  the  monomial  common  to  the  terms  of 
the  given  expression. 

This  monomial  is  one  factor,  and  the  expression  obtained  by 
dividing  the  given  expression  by  it  is  the  other  factor. 


TRINOMIAL  EXPRESSIONS  101 

Illustrations : 

1.  x8  -  x2  +  3x  =  x(x2  -  x  +  3). 

2.  3  m4  +  9  ro8  -  12  ro2  +  15  ro  =  3  ro  (ro8  +  3  ro2  -  4  ro  +  5). 

3.  3  a262  -  6  a8*)2  +  6  a864  -  9  a465  =  3  a2b2  (1  -  2  a  +  2  a&2  -  3  a26«). 

Exercise  25 

Factor  orally : 

1.  5a- 106  +  15c.  7.  5x —  20xy +  35xz. 

2.  8m  —  16n-f-24a;.  8.  a&  +  3ac+7aa\ 

3.  12  a  +  18  y  +  24  2.  9.  5  m  —  20  ran  —  40  raz. 

4.  5 c  +  10 a"  +  15 k.  10.  a4  +  a3-a2  +  a. 

5.  15x  +  30y  +  4:5z.  n.  ic8  —  a^  —  a;4  —  ic2. 

6.  14ra-21n  +  14.  12.  x2  -  x5  +  x*  -  x9. 
Write  the  factors  of: 

13.  5  <*b-20  cW  +  35  car5. 

14.  17  a4»+  51  a3ar2-34aV-85aa!4. 

15.  27  ra3n  +  18  ra2n  —  9  ran2  —  36  ran3. 

16.  cPa?y  —  a2xy  +  aa^2  —  aary. 

17.  asm3n3  —  a3m4n2  —  cPmhi3  —  a3m4n5. 

18.  14  a46  +  21  a362  -  35  a263  -  42  a&4- 

19.  3a?z-6<x&  +  9a&-12a&  +  15x*. 

20.  a3™,  —  3  a2cm  —  a62ra  —  ara3  +  2  ara2  4-  ara. 

TRINOMIAL  EXPRESSIONS 

(a)  The  Perfect  Trinomial  Square 

Type  Form  ...  x2  ±  2  xy  +/. 
The  Origin:  If  a  binomial  (x±y)  is  multiplied  by  itself,  or 


squared,  we  have : 


by  Art.  Ill :  (x  +  y)2  =  x2  +  2  xy  +  t/2. 

or  by  Art.  112  :  (x  -  y)2  =  x2  -  2  xy  +  y2. 


102  FACTORING 

In  each  product : 

The  first  term  is  a  perfect  square ;  ] 
The  third  term  is  a  perfect  square.  J    Both  are  +* 
The  second  term  is  twice  the  product  of  1    Its  sign  may  be 
the  square  roots  of  the  square  terms,  j      either  +  or — . 
These  are  the  conditions  of  a  perfect  trinomial  square.    In  like  manner : 
c2-  18c  +  81,       4a2_i2a&  +  962,        16 xY  -  40 x2y2z  +  25 z* 

are  all  perfect  trinomial  squares,  for  each  trinomial  has  two  positive 
square  terms,  with  a  second  term  that  is  twice  the  product  of  the  square 
roots  of  the  square  terms. 

Hence,  to  factor  a  perfect  trinomial  square  : 

130.  If  necessary ;  arrange  the  terms  of  the  given  expression  in 
order. 

Take  the  square  roots  of  the  first  and  last  terms,  and  connect 
them  with  the  sign  of  the  second  term. 

The  resulting  binomial  is  one  of  the  two  required  equal  factors. 

Illustrations : 

1.  x2  +  16  x  +  64  =  (x  +  8)  (x  +  8)  =  (as  +  8)2. 

2.  9  m2  -  12  mn  +  4  n2  =  (3  m  -  2  n)  (3  m  -  2  n)  =  (3  m  -  2  w)2. 

3.  x6  -  26  «8  +  169  =  (x3  -  13)  (x3  -  13)  =  (x3  -  13)2. 

Exercise  26 

Factor  orally : 

1.  ^  +  6^  +  9.      5.   m2-r-14m  +  49.         9.   raV '-22 mn  +  121. 

2.  x2  +  8^+16.     6.    c2- 18c +  81.  10.  64  +  16z  +  x2. 

3.  ^-6^  +  9.      7.    w2  +  20n  +  100.  11.  81+m2-18m. 

4.  7?  -  8  x  + 16.     8.   a2b2  - 12  ab  +  36.  12.  144  -  24  xy  +  x2y2. 
Write  the  factors  of : 

13.  9c2-6cd  +  d2.  16.   4(P+12dn  +  9na. 

14.  4P-4fcm  +  m2.  17.    16 c2  +  24 ex  +  9 ar2. 

15.  8a2/  +  2/2  +  16n2.  18.   25  p2  -  30#  +  9. 


TRINOMIAL   EXPRESSIONS  103 

19.  36^  +  25m2  -60cm.       *    22.   64 x4  -  80 x*y2  +  25 y4. 

20.  72  cc?  +  16^  +  81  d2.  23.   49  a2b2  +  140  a&c  f  100  c2. 

21.  36^  +  84^  +  49.  24.   81aV-198a2a*/  +  121^. 

(6)  The   Trinomial  whose   Highest   Power   has  the  Coeffi- 
cient Unity 

Type  Form  •••  x2  +  (c  +  </)*  +  cfl^ 

The  Origin :  If  two  binomials,  (x  +  c)  and  (x  +  d),  are  multi- 
plied, we  have  (Art.  115)  : 

(x  +  c)(x  -{•  d)=  x(x  +  d)  +  c  (x  +  d) 

=  x2  +  dx  +  cx  +  cd 

=  x2  +  (c  +  d)  x  +  cd. 
In  the  resulting  product  : 

The  first  term  x2  =  the  product  of  the  given  first  terms. 

The  second  term  (c+d)x  =  the  product  of  the  given  first  term  by  the  alge- 
braic sum  of  the  given  second  terms. 
The  third  term  cd  =  the  product  of  the  given  second  terms. 

Hence,  the  coefficient  of  the  second  term  of  the  product  is  the 
sum  of  the  given  second  terms. 
Illustrations  : 

1.  x2  +  8  a?  +  15  =  (x  +  ?)  (x  +  ?). 

We  require  the  two  factors  of  +  15  whose  sum  is  +  8  :    +3  and  +  5. 
Therefore,  x2  +  8  x  +  15  =  (x  +  3)  (x  +  5).    Result. 

2.  x2  -  8  x  +  15  =  (x  -  ?)  (x  -  ?). 

We  require  the  two  factors  of  +  15  whose  sum  is  —  8  :    —  3  and  —  5. 
Therefore,  x2  -  8  x  +  15  =  (x  -  3)  (x  -  5).     Result. 

3.  x2  +  2a?  -  15  =  (as  +  ?)  (x  -  ?). 

In  this  expression  the  sign  of  15  is  — ,  hence  the  signs  of  its  factors 
are  unlike.     The  sign  of  2  is  +,  hence  the  greater  factor  of  —  15  is  +. 
We  require  the  two  factors  of  —  1 5  whose  sum  is  +  2  :    +5  and  —  3. 
Therefore,  x2  +  2  x  -  15  =  (x  +  6)  (x  -  3).     Result. 


104  FACTORING 

4.  x2-2x~  15  =  (x-?)  0+?). 

In  this  expression  the  sign  of  15  is  — ,  hence  the  signs  of  its  factors  are 
unlike.    The  sign  of  2  is  — ,  hence  the  greater  factor  of  —  15  is  — . 
Therefore,  x2  -  2  x  -  15  =  (x  -  5)  (x  +  3) .    Result. 

From  the  illustrations  we  make  the  following  important 
conclusions : 

131.  (1)  The  first  aid  to  factoring  such  expressions  is  the  sign 
of  the  third  term. 

(2)  If  that  sigyi  is  +,  the  signs  of  the  second  terms  in  the 
required  factors  are  like ;  but  if  that  sign  is  — ,  the  signs  of  the 
second  terms  of  the  factors  are  unlike. 

(3)  The  sign  of  the  given  second  term  is  the  same  as  that  of  the 
greater  factor  of  the  given  third  term. 

Exercise  27 

Give  orally  the  factors  of  the  three  groups,  a,  b,  and  c. 

(a)  The  third  term  + .  The  second  term  -f- .  The  sign  of 
the  last  term  of  each  factor  +• 

Illustration :  x2  +  10  x  +  24  =  (x  +  4)  (x  +  6). 

1.  a?  +  7x  +  12.  5.    d2  +  20d  +  36. 

2.  m2  +  8m-}-l5.  6.   f  + 19 y  +  48. 

3.  c2  +  l2c  +  20.  7.    c2  +  31c  +  58. 

4.  a2+12a  +  32.  8.  p2  +  37p+70. 

(b)  The  third  term  +.  The  second  term  — .  The  sign  of 
the  last  term  of  each  factor  — . 

Illustration  :  x2  -  10  x  +  24  =  (x  -  4)  (x  -  6). 

9.  a2_8a  +  12.  13.  x2-18x  +  32. 

10.  ^-9a;  +  18.  14.  c2-16c  +  39. 

11.  c2-9c  +  14.  15.  ?i2-ll7i  +  24. 

12.  m2-10m  +  21.  16.  ?/2 - 14 */ +  24. 


TRINOMIAL  EXPRESSIONS  105 

(c)  The  third  term  — .  The  second  term  either  +  or  — . 
The  signs  of  the  last  terms  of  the  factors  unlike,  the  greater 
last  term  having  the  same  sign  as  the  second  term  of  the  given 
trinomial. 

Illustrations :    x2  +  2  x  -  24  =  (as  +  6)  (as  -  4). 
x2  -  2  x  -  24  =  (x  -  6)  (x  +  4). 

17.  tf  +  x-20.  21.  c2-15c-34 

18.  ra2-3ra-10.  22.  a2-9a-70. 

19.  C2_5C_14.  23.  z2  +  13z-48. 

20.  a^  +  8a;-20.  24.  a?-6x-72. 

(d)  Write  the  factors  of : 

25.  m2- 15  m -54.  34.  ra2n2  -  8  ran  -  84. 

26.  ^-c-m  35.  x2y2-72-xy. 

27.  ?/2-  11 2/  -26.  36.  28  - 16  xz  +  arV. 

28.  110-53C-C2.  37.  ah2 -15  az- 76. 

29.  afy2-25a;2/  +  46.  38.  96  +  28  xyz  +  afyV. 

30.  n2-9n-112.  39.  c*  +  33c?-70. 

31.  2  a; -120 -far2.  40.  63  a262+a464-130. 

32.  &2  +  13  6-140."  41.  ra2n2  -  ran  -  210. 

33.  d2-6d -135.  42.  aW -  7  a262c - 144. 

43.  a2c?x2  +  5acx-36. 

44.  45  +  4  ran2  —  ra2n4. 

45.  33+8cti-cftft 

46.  54-15ar5-a* 

47.  ac2d-aW  +  12. 

48.  2  mux  —  mWx2  + 143. 

49.  380-cc?V-<fW, 


106  FACTORING 

(c)  The  Trinomial  whose  Highest  Power  has  a  Coefficient 
Greater  than  Unity 

Type  Form  •  •  •  acx2  4-  (ad  4-  be) x  4-  bd. 

The  Origin:  If  two  binomials,  (ax  +  b)  and  (cx  +  d),  are 
multiplied,  we  have  (Art.  116) : 

(ax  +  b)  (ex  4-  a")  =  ax  (ex  4-  a")  -f  6  (ex  +  d) 
=  acx2  4-  adx  +  6cx  +  bd 
=  acx2  +  (ad  4-  6c)  x  +  bd. 

Now  the  coefficient  of  x  in  the  result  (ad -{-be)  is  made 
up  of  the  coefficients  that,  multiplied,  would  produce  abed. 
Therefore, 

132.  We  require  the  two  factors  of  abed  that,  added,  will 
produce  ad  +  be. 

The  application  of  the  principle  in  practice  will  be  readily 
understood  from  the  following  illustrations : 

1.  Factor  6x*  +  25  x  +  14. 
6  x  14  =  84. 

Required  the  factors  of  84  that,  added,  equal  25. 

By  trial  we  find  them  to  be  4  and  21. 

Therefore,       6  x2  +  25x  +  14  =  6  x2  +  4  x  +  21  x  +  14 

=  (6x24-4x)  +  (21x  +  14) 
=  2x(3x  +  2)+7(3x+2) 
=  (2  x  +  7)  (3  x  +  2) .     Result. 

2.  Factor  14  a2  +  31  a  -10. 

14  x  -10=  -140. 

Required  the  factors  of  —140  that,  added,  equal  31. 

By  trial  they  are  found  to  be  4-35  and  —4. 

Therefore,     14  a2  4-  31  a  -  10  =  14  a2  +  35  a  -  4  a  -  10 

=  (14a24-35a)-(4a4-10) 
=  7a(2a4-5)-2(2a4-5) 
=  (7  a  -  2)  (2  a  +  5) .    Result. 

Several  excellent  methods  for  factoring  a  trinomial  of  this  type  form 
might  be  given,  but  to  understand  and  apply  accurately  one  method  is  a 
better  plan  for  the  beginner. 


BINOMIAL  EXPRESSIONS  107 

Exercise  28 

Write  the  factors  of : 

1.  4m2  +  8m  +  3.  14.  9m2  +  23m  +  10. 

2.  2c2  +  3c  +  l.    .  15.  3z2  +  52-22. 

3.  6z2-7z  +  2.  16.  6a2  +  7a&-5&2. 

4.  6d2-lld  +  4.  17.  12m2-23m7i-f-10w2. 

5.  9?i2r-9n  +  2.  18.  10c?-3cd-18d2. 

6.  6  2/2  — 13  2/  -h  6.  19.  6  z2  -  31  zz  +  35  z2. 

7.  6c2  +  17c  +  12.  20.  16c2  +  18cd-9d2. 

8.  8a2-2a-3.  21.  10  a2  +  ax  -  24=  a2. 

9.  9m2  +  21m  +  10.  22.  9 v2 - 15 vx - 50 x2. 

10.  6z2-7z-5.  23.  7c2-50cz  +  7z2. 

11.  lOrf-Ux  +  S.  24.  20a2-27am-14m2. 

12.  10  a2 +  <e -3.  25.  10  z2-zm- 21m2. 

13.  6z2-z-12.  26.  Sn2-10nx-25x2. 

27.   20a2-37az-18zV 

BINOMIAL  EXPRESSIONS 

(a)  The  Difference  of  Two  Squares 

Type  Form  •••  x2  —/*. 

The  Origin:     If  two  binomials,  (x  +  y)  and  (x—y),  are  mul- 
tiplied, we  have  (Art.  113): 

(x  +  y)(x-y)=x*-y*. 

Therefore,  to  factor  the  difference  of  two  squares : 

133.   Extract  the  square  roots  of  the  given  square  terms. 
One  factor  is  the  sum  of  these  square  roots  ;  the  other  factor, 
their  difference. 


108  FACTORING 


Illustrations : 

1.  4a2-25a2  =  (2a  +  5z)(2a 

2.  25a4  -  9s6  =  (5a2  +  3s3)  (5 

3.  16  a4  -  81   =  (4  a2  +  9)  (4  a2 

=  (4a2 +  9)  (2a 

-5x). 
a2-3s»). 

-9) 
+  3)  (2a -3). 

Note  that  the  second  factor  of  (3)  can  be  refactored  into 
two  other  factors. 

Exercise  29 
Factor  orally : 

1.   tf-y2.      6.  d2  -16.     ll.   9^-25. 

16.   4tx*-25y2. 

2.   a?-z2.       7.   a2 -25.      12.    16  m2 -9. 

17.   16ra2-49n2. 

3.   c2  -1.        8.  m2-49.     13.    4w2-25. 

18.   36  a2- 25  z2. 

4.   a2 -4.        9.    w2-81.      14.    9a2-64. 

19.   49  n2- 100  #2. 

5.   a2-9.      10.   4c2-9.      15.   a?-9f. 

20.   121a262-144c2. 

Write  the  factors  of : 

21.   aA  —  xA.                    25.    a8  —  a8. 

29.   ra4-926. 

22.   c4-16d4.                26.   c10-16. 

30.   c12-64m4. 

23.    64  x4-  if,                27.    c8-256. 

31.   36aV-25y«. 

24.   81m4-16w4.         28.   <c16-l. 

32.   49  m10  -4n1V4. 

(6)  The  Difference  of  Two  Cubes 

Type  Form  •••  x*—f. 

The  Origin :  Since  the  product  of  a  divisor  by  a  quotient 
equals  the  corresponding  dividend,  we  have,  from  Art.  119,  an 
expression  (x*  —  if)  equal  to  the  product  of  the  expressions 

(x  -  y)  and  (x2  +  xy  +  y2). 

Therefore,  for  the  factors  of  the  difference  of  two  cubes : 

134.  One  factor  is  the  difference  of  the  cube  roots  of  the 
quantities.  The  other  factor  is  the  sum  of  the  squares  of  the  cube 
roots  of  the  quantities  plus  their  product. 


BINOMIAL   EXPRESSIONS  109 

Illustrations : 

1.  c3-8  =  (c-2)(c2  +  2c-f  4). 

2.  27  ra3  -  64  =  (3  w  -  4)  (9  ro2  +  12  m  +  16). 

3.  64  s6  -  125  y%3  =  (4  x2  -  5  yz)  (16  a*  +  20  afys  +  25  y2*2). 


Factor  orally : 

Exercise  30 

1.   c3-d3. 

5. 

m3-8. 

9. 

27  -s3. 

2.  a3  — m3. 

6. 

a3 -27. 

10. 

64-a3. 

3.   w3-^. 

7. 

f-64:. 

11. 

cW-64. 

4.   c3-!. 

8. 

c?-125. 

12. 

a3m3- 125 

Write  the  factors  of : 

13.  8^-27.  17.   8  m3  -343.  21.  125m9-l. 

14.  27 -125  m3.      18.   8  c6 -27.  22.  27  m6n3  -  64  x9. 

15.  125 x3-  64  f.    19.    512  ra3- a6.  23.  125 x« -  729 z12. 

16.  216 a3 -ay.     20.    125  ay  -  64  z6.  24.  729 m12- 1000 ny. 

(c)  The  Sum  of  Two  Cubes 

»  Type  Form  •••  **+/. 

The  Origin :   As  in  the   preceding   case,   we  refer   to  the 
principle  of  division  by  which  (aP  +  y3)  is  shown  to  contain 

I(x  +  y),  the  quotient  being  (zP  —  xy  +  y2)  (Art.  120). 
Therefore,  for  the  factors  of  the  sum  of  two  cubes : 
135.    One  factor  is  the  sum  of  the  cube  roots  of  the  quantities. 
The  other  factor  is  the  sum  of  the  squares  of  the  cube  roots  of  the 
quantities  minus  their  product. 

Illustrations : 

1.  c3  +  27  =  (c+3)(c2-3c+9). 

2.  8+125c3  =  (2  +  5c)(4-10c  +  25c2). 

3.  27  z6  +  64  w9  =  (3  a2  +  4  n3)  (9  x*  - 12  se2w3  +  16  n«). 


110  FACTORING 

Exercise  31 

Factor  orally : 

1.  a8  +C8.         4.   n8  +  l.  7.  m3  +  125.  10.  64  +  afy3. 

2.  rrt  +  tf.         5.    a^  +  8.  8.  2/3  +  216.  11.  m3n3  +  343. 

3.  ds  +tf.         6.   z3  +  27.  9.  27  +  a363.  12.  2/V  +  512. 
Write  the  factors  of : 

13.  8  a3  +  27&3.       17.   125  a6 +  1.  21.  512  + 125  aP. 

14.  64  a3 +125.         18.    64  a6 +  27^.       22.  729  +  64  aft/3. 

15.  27ra3  +  64n3.       19.   xY  +  125z\       23.  1000  v?  +  27  yl\ 

16.  125  tt3  + 216  ar5.     20.   ay  +  216z3.       24.  1728  a15  + 1331  yu. 

EXPRESSIONS  OF  FOUR  OR  MORE  TERMS  FACTORED  BY  GROUPING 

(a)  The  Grouping  of  Terms  to   show  a  Common  Polynomial 

Factor 

Type  Form  •  •  •  ax  +  ay  +  bx  -f  by. 

The  Origin:  If  any  binomial,  (%  +  y),  is  multiplied  by  a 
binomial  having  dissimilar  terms,  (a  +  6),  we  have  (Art.  55)  : 

(a  +  b)  (x  +  y)  =  a(x  +  y)  +  b  (x  +  y) 
=  ax  +  ay  +  bx  +  &y. 

In  the  resulting  product  note  that  a  is  common  to  the  first  two  terms, 
and  that  b  is  common  to  the  last  two  terms.  Note,  also,  that  dividing 
the  first  two  terms  by  a,  and  the  last  two  terms  by  6,  gives  the  same 
quotient,  (x+y). 

Therefore  :     (x  +y)  is  a  common  polynomial  factor. 

And  from  a  (x  4-  y)  +  b  (x  +  y)  we  obtain  by  adding  the  coefficients 
of  the  common  factor  :     (a  +  6)  (x  +  y) ,  the  factors. 

Illustrations : 

1.   ac  +  bc+  ad+bd  =  (ac  +  6c)  +  (ad  +  bd) 

=  c(a  +  b)  +  d(a  +  b) 
=  (c  +  d)(a  +  b). 


EXPRESSIONS  OF  FOUR   OR  MORE  TERMS  111 

2.  2  c2  -  6  c  +  cd  -  8  d  =  (2  c2  -  6  c)  +  (cd  -  3  d) 
=  2c(c-3)  +  d(c-3) 
=  (2c  +  d)(c-3). 

3.  a86  -  14  -  7  a2  +  2  a&  =  a35  +  2  a&  -  7  a2  -  14 

=  (a36  +  2  aft)  -  (7  a2  +  14) 

=  a&(a2+2)-7(a2  +  2) 
=  (a6-7)(a2  +  2). 

136.   Note  that  the  proper  arrangement  of  an  expression  is 
I   first  necessary.     We   are  assisted  in  grouping  the  terms  by 
noting  that  terms  bearing  to  each   other  the  same  relation  are 
grouped  together. 

With  expressions  of  more  than  four  terms  the  principle  is 
unchanged. 

Exercise  32 
Write  the  factors  of : 

1.  am  +  an  +mx  +  nx.  10.   abxy  —  cxy—cz  +  abz. 

2.  am  +  an  —  cm  —  en.  11.   x2  +  bx  +  ax  -f-  ab. 

3.  ac  —  ad  —  be  +  bd.  12.   y2  —  my  +  2y  —  2m. 

4.  ax  +  ay  -f-  x  +  y.  13.  .x3  +  ax2  +  x  -f  a. 

5.  ax  —  az  —  x  +  z.  14.   z3  +  2  2  —  3  z2  —  6. 

6.  aai-f  3  a +  2  a; +  6.  15>   a6  +  5  a3  +  10  +  2  a2. 

7.  a2/-4a  +  52/-20.  16    x4  +  i41_2x-7  3?. 

8.  afoc  —  2  a&  -f  ca>  —  2  c.  17.    —  arfcc  -f  3  d  —  3  ex  -f-  aca^. 

9.  2rana;-5a;-6mw+l5.      18.    3  d-  10  d2-15  +  2  d*. 

(b)  The  Grouping  of   Terms  to   form   the  Difference  of 
Two  Squares 

Type  Form  ...  x2  +  2  xy  +/  -  z2. 
The  Origin :  By  multiplication, 

(x  +  y  +  z)(x  +  y-z)  =  [(a  +  y)  +  *][(*+»  -  z] 
=  (X  +  t/)2  -  z2 
=  a;2  +  2  a;y  +  y2  -  s2. 


112  FACTORING 

The  product  is  the  difference  of  a  trinomial  square  and  a 
monomial  square.  Therefore,  an  expression  of  this  type  is 
factored  by  grouping  three  of  the  terms  that  will  together 
form  a  perfect  trinomial  square ;  the  fourth  term  being  a  per- 
fect monomial  square.  The  result  is  a  difference  of  two 
squares. 

Illustrations : 

1.  a2  +  2  ab  +  b2  -  c2  =  (a2  +  2  ab  +  62)  -  c2 

st  (a  +  by  -  c2 

=  (a  +  b  +  c)(a  +  b-c). 

2.  x2  -  y2  -  2  2/0  -  z2  =  x2  _  (y2  +  2  ^  +  s*) 

=  x2  -  (y  +  zY 

«[•+ ft +!«)K^- &■+"#)] 

=  (»  +  y +  «)(&-?-«)• 

3.  a2  -  6  ax  +  9  at2  -  4  m2  -  12  m  -  9 

=  (a2  -  6  ax  +  9  x2)  -  (4  m*  +  12  m  +  9) 

=  (a-3x)2-  (2m  +  3)2 

=  [(a  -  3  x)  +  (2  m  +  8>]  [(a  -3  x)  -  (2  m  -f  3)] 

=  (a  -  3  x  4-  2  m-\-  3) (a  -  3  x  -  2  m  -  3). 

The  process  consists  mainly  in  finding  three  terms  that,  when 
grouped,  form  a  perfect  trinomial  square.  The  key  to  the  group- 
ing is  the  given  term  that  is  not  a  perfect  monomial  square.  Or, 
considering  the  given  square  terms,  we  may  state : 

137.  (1)  When  only  one  given  square  term  is  plus,  it  is  written 
first,  and  the  other  three  terms  are  inclosed  in  a  negative  paren- 
thesis. 

(2)  When  only  one  given  square  term'  is  minus,  it  is  written 
last,  and  the  other  three  terms  are  written  first  in  a  positive 
parenthesis. 

Exercise  33 

Write  the  factors  of : 

1.  a2  +  2ax  +  a?-m2.  3.   c2  +  d2-y2-2 cd. 

2.  tf-2yz  +  z*-4:.  4.   l-c2-2cd-tf. 


REPEATED  FACTORING  113 

5.  v?-m2-l  +  2m.  10.  c2 - 10 ex  +  25 x2 - 49 m2. 

6.  4c2_<f_j_1_4c>  X1>  a^-4a;2/-9»22/24-4?/2. 

7.  9-m2-2m2/-2/2.  12.  2/2  +  16z2-  16  +  8  yz. 

8.  4c2  +  c4-4c3-4.  13.  ary-a2&2  +  16a&-64. 

9.  12x  +  ±x2-9y2  +  9.  14.  490^  +  10  c2^2-.^4- 25. 

15.  81 -100  x*y* +  60  x4y2z -9  z2. 

16.  4iB2-4»  +  l-9m24-6wi?i-7*2. 

17.  c?  —  a?  +  x2  —  y2  —  2cx  —  2  ay. 

18.  c2-d2-x2  +  m2  —  2cm  +  2dx. 

19.  36c44-l-49ic6-92/2-12c2-42ofly. 

REPEATED  FACTORING 

138.   Any  process  in  factoring  may  result  in   factors  that 

|  may  still  be  resolved  into  other  factors.     The  review  examples 

following  frequently  combine  two  or  more  types  already  con- 

\  sidered.     The   following   hints   on  factoring   in   general  will 

i  assist. 

1.  Remove  monomial  factors  common  to  all  terms. 

2.  What  is  the  number  of  terms  in  the  expression  to  be  fac- 
tored ? 

(a)  If  two  terms,  which  of  the  three  types  ? 

(b)  If  three  terms,  which  of  the  three  types  ? 

(c)  If  four  terms,  how  shall  they  be  grouped  ? 

3.  Continue  the  processes  until  the  resulting  factors  are  prime. 

Illustrations : 

1.  Sa7-3ax*  =  3a(a?-x6) 

=  3  a  (a3  +  x8)  (a8  -  x3) 

=  3  a  (a  +  x)  (a2  -  ax  +  x2)  (a  -  x)  (a2  +  ax  +  x2). 

2.  12x5-75x8  +  108x=3x(4x4-25x2  +  36) 

=  3  x  (4  x2  -  9)  (x2  -  4) 

=  3x(2x  +  3)(2x-3)(x  +  2)(x-2). 


SOM.    EL.    ALG. 8 


J 


114  FACTORING 

3.  8  a8x  -  2  ax  +  8  a5x  -  2  a4x  =  8  a8x  +  8  a6x  -  2  a4x  -  2  as 

=  2  ax  (4  a7+  4  a4  -  a8  - 1)      • 
=  2  ax  [(4  a7  +  4  a4)  -  (a8  + 1)] 
=  2  ax  [4  a4  (a3  +  1)  -  (a8  +  1)] 
=  2ax(4a4-l)(a3+l) 
=  2ax(2a2+l)(2a2-l)(a+l)(a2-a  +  l). 

MISCELLANEOUS   FACTORING 

Exercise  34 

Write  the  factors  of : 

1.  5a?  +  25x  +  20.  21    75  a3 -90 a2 -{-27  a. 

2.  5  0^-45  a3.  22.  6a2x+9abx-8a2y-12aby. 

3.  8  a2  -  24  a  + 18.  23.  5  m3  + 135. 

4.  a4-8a.  24.  15  c3  +33  c2  +  6c. 

5.  2am  +  2ma;-4a-4aj.  25.  242  ar*- 98  a. 

6.  2-2c2-4cd-2d2.  26.  12  m4  + 10 m3-8  m2. 

7.  3  m4  +  81  m.  27.  160  a,-5  +  20  x2. 

8.  3a2-9a-84.  28.  3  ar5  - 15  a3  + 12  a;. 

9.  2c5- 128c2.  29.  98  ar>  + 18  a*/2- 84  afy. 

10.  16  m V  +  8  mnx  +  x2.  30.  m4- 2  m3+ 8m- 16. 

11.  15  -  60  xy  +  60  x*y2.  31.  15  a3  -  25  a2a  - 10  ax2. 

12.  7  aV  - 175  ax.  32.  81  afy4  -  3  a#. 

13.  c4+c2-3c3-3c.  33.  4cd  +  2c2d2  +  2-2arJ. 

14.  Ux*-S2x2-12x.  34.  250c4-16c. 

15.  49m3-84m2  +  36m.  35.  8mV+4m¥-112mw. 

16.  4o3-36arJ  +  56a;.  36.  5  c5- 10  c3- 315  c. 

17.  7  a7-  7x.  37.  21  ar>  +  77  a2 -140  a;. 

18.  4  a3  +  16  a2  + 15  a.  38.  512  -  32  m4n\ 

19.  8a2-18c2  +  24a&  +  1862.    39.  27^+215^-8. 

20.  24^-55^-24^  40.  3  x»  -51  a^  +  48  x. 


SUPPLEMENTARY  FACTORING  115 

41.  686  x*  -  2  xtf.  45.  242  a4  -  748  a2  +  578. 

42.  8m5  +  8m2-18m3-18.  46.  2x5+64y2-$xstf-16x2. 

43.  8a3  +  2a5-8a4-2a.  47.  64a6  +  729a2. 

44.  32  a4-  2ajy.  *»•  8  a*  - 10  a8  -  432  a. 

SUPPLEMENTARY  FACTORING 

(a)  Compound  Expressions  in  Binomial  and  in  Trinomial  Forms 

Comparative  illustrations  with  corresponding  processes  : 

1.  z2  +  12z+35  =  (x  +  5)(z  +  7). 

Similarly, 

(a  _  &)2  +  12  (a  -  6)  +  35  =  (a  -  6  +  5)(a  -6  +  7). 

2.  3£2  +  10zy  +  32/2  =  (3s  +  y)(z  +  3y). 

Similarly, 

3  (a  -  b)2  +  10  (a  -  6)  (c  -  d)  +  3  (c  -  d)2 

=  [8(a-&)  +  (c-<I)][(a-&)+3(c-tf)] 
=  (3a-3  6  +  c-d)(a-6  +  3c-3d). 

3.  a2-9a2  =  (a  +  3z)(a-3x). 

Similarly, 

(«  +  2)2-9(x-l)2  =  [(a;+2)+3(x-l)][(x  +  2)-3(a;-l)] 

=  (z  +  2  +  3o;-3)(a;  +  2-3a;  +  3) 

=  (4x-l)(-2x-f  5) 

=  _(4z-l)(2x-5). 

4.  a8  +  8  =  (a  +  2)(a2-2a  +  4). 

Similarly, 

(a  -  6)8  4-  8  =  [(a  -  6)  +  2] [(a  -  6)2  -  2  (a  -  6)  +  4] 

s  (a-6  +  2)(a2-2a6  +  62-2a  +  26  +  4). 

139.  If  similar  monomial  terms  occur  in  the  different  compound 
terms  of  an  expression,  the  factors  can  usually  be  simplified  by 
collecting  like  terms. 


116  '  FACTORING 

Exercise  35 

Find  and  simplify  the  factors  of : 

1.  16(a+l)2-9a^.  6.  (x2  - 1)2 - 9 (x  +  2)2. 

2.  49(c-l)2-4.  7.  27(a-l)3  +  a3. 

3.  (a  +  %)2  + 10  (a  +  x)  +  24.         8.  2  (x  +  l)2  + 11(*+1)  +  12. 

4.  (m  +  2)2  -  15  (m+2)  +  56.       9.  (2  m-  l)4  -  16  (2  m  + 1)4. 

5.  (2/_3)2_7(^_3)_30.         io.  8(a-2)3-27(a  +  l)3. 

11.  6a2-12aaj  +  6^-5a  +  5a;4-l. 

12.  3(s+  l)2  +  7(>2-l)-6(a-l)2.     . 

(6)  The  Difference   of   Two  Squares   obtained  by  Addition 
and  Subtraction  of  a  Monomial  Perfect  Square 

140.  No  general  statement  of  this  process  can  be  given  in  a 
simple  form. 

Illustration : 

Factor  9^+6^  +  49. 

The  first  and  third  terms  are  positive  perfect  squares. 
Hence,  if  the  middle  term  were  twice  the  product  of  their  square 
roots,  the  expression  would  be  a  perfect  trinomial  square. 

For  a  perfect  trinomial  square  the  middle  term  should 
be  2  (3  2^(7)  =  42  x2.  Adding  +  36  x2  to  the  expression,  we 
obtain  the  perfect  trinomial  square  required,  and  subtracting 
the  same  square,  +  36  x2,  the  expression  is  unchanged  in  value 
but  is  now  the  difference  of  two  perfect  squares.     Therefore : 

9z4  +  6z2  +  49  =    9a4+   6  x2  +49 

+  36  x2  -  36  x2 

=    9  &  +  42  x2  +  49  -  36  x2 

=  (3x2  +  7)2-36s2 

=  (Sx2  +  7+6x)(3x2  +  7-6x). 

It  is  important  to  note  that  a  positive  perfect  square  only  can 
be  added. 


SUPPLEMENTARY  FACTORING  117 

Exercise  36 

Write  the  factors  of: 

1.  a^  +  ^  +  l.  7.  25^-51^  +  25. 

2.  a4  +  3a2  +  4.  8.  81  m4  +  45  ra2  +  49. 

3.  n4-7n»  +  l.  9.  36c4-61c2m2  +  25m4. 

4.  (j* -28  c2  +  16.  10.  64  a4 +  79^ +  100  a4. 

5.  9c4  +  5cV  +  25a;4.  11.  64  m4  +  76  m2n2  +  49  n4. 

6.  9  d4-  55  d?x2  +  25  a;4.  12.  81  x4 - 169  zV  +  64  z\ 

(c)  The  Sum  and  the  Difference  of  Equal  Odd  Powers 

By  actual  division : 

<z5  +  x5 


a  +  x 
a6  — x5 


a*  —  asx  +  ah?  —  ax3  +  x4. 
=  a4  +  a3x  +  a>2x2  +  ax3  +  x4. 


a  — x 
Hence,  for  the  factors : 

a5  +  x5  =  (a  +  x)  (a4  -  a3x  +  a2x2  -  ax3  +  x4). 
a5  -  x6  =  (a  -  x)  (a4  +  a3x  +  a2x2  +  ax3  +  x4). 

By  Art.  121,  the  factors  of  any  similar  cases  may  be  found. 

In  general,  therefore : 

141.  (1)  One  of  the  factors  of  the  sum  of  equal  odd  powers  is 
the  sum  of  the  quantities.  The  other  factor  is  the  quotient  obtained 
by  using  the  binomial  as  a  divisor. 

(2)  One  of  the  factors  of  the  difference  of  equal  odd  powers  is 
the  difference  of  the  quantities.  The  other  factor  is  the  quotient 
obtained  by  using  the  binomial  as  a  divisor. 


Write  the  factors  of : 

Exercise  37 

- 

1.   ra5-l. 

5. 

ra5  +  n5. 

9. 

32(^+243. 

2.    tf  +  x*. 

6. 

32  -  n5. 

10. 

a¥-32. 

3.   m7  —  n7. 

7. 

128 +c7. 

11. 

m11  +  w11. 

4.   a7  +  x7.      ■ 

8. 

a?Y- 343. 

12. 

aiociD_ajy. 

118  FACTORING 

MISCELLANEOUS  FACTORING 

Exercise  38 

1.  3a*- 54^  +  243a.  14.  72 m8  +  94 m4  + 128. 

2.  (c3  +  8)  +  2c(c  +  2).  15.  2a2(a+2)-5a2-8a+4. 

3.  4a2  +  9ar>-*-(12aa;  +  16).  16.  (c2  +  a2)2  -  4  c'x2. 

4.  5(c?  +  l)-15c-15.  17.  (4ar5+32)-5a2-17a-14. 

5.  (m2-12)2-m2.  18.  98  a5  -  16ajV  +  8aw*. 

6.  (a  +  2)2-7(a  +  2)  +  12.  19.  m6-3m4  +  3m2- 1. 

7.  4  +  2  a? —  (am +  05) am.  20.  (a^  +  a- l)2-(a2-a-l)2. 

8.  3(a  +  l)2-19(a  +  l)+6.  21.  2(ar3-l)  +  7(a2-l). 

9.  (m2-12)2-(m2-6)2.  22.  m4-9a4  +  m2  +  3a2. 

10.  2  a6 +  38  a3 -432.  23.  12m2-m(w-l)-(n-l)2. 

11.  (a2  +  4)3-125a3.  24.  2  d10  - 1024  a\ 

12.  (m+fl?)2— 7— 3(m+a;+l).  25.  cx+my—mx—cz—cy-\-mz. 

13.  (c2  +  4)3-16c4-64c2.  26.  4  mV  -  (m2  +  n2  - 1)2. 

27.  (c  +  d)(m2-l)-(m  +  l)(c2-c?2). 

28.  1262-18ma  +  12a6-3a2  +  3a2-27m2. 

29.  7m2  —  7x2  +  7n2  —  14(az  —  mri)  —  7  z2. 

30.  (m  —  l)(a2  —  asc)  +  (l  —  m)(ax  —  x2). 

31.  (a-3)3-2(a-3)2-15(a;-3). 

32.  (a2  +  4a  +  3)2-23(a2  +  4a  +  3)  +  120. 

33.  (x-2)(x-3)(x-4:)-(x-2)  +  (x-2)(x-3). 

34.  a?b  +  b2c  +  ac2  -  a?c  -  a&2  -  6c2. 

35.  (3a  +  2)(9a2  +  2a;  +  12)~(27aj3  +  8). 

36.  8(a-a)2  +  5a2-5a2-3(a  +  a)2. 

37.  (c-l)(c2-9)-(c  +  5)(c- 3)- 3(^  +  9  0. 

38.  (x  +  2y)2-3(x  +  2y  +  l)-W. 

39.  m2(x  —  1)  —  2  mx  —  m  +  x2  (m  —  1)  —  x. 

40.  ^(a-5)2  +  2a(a2-a;-20)  +  (a;  +  4)2. 


CHAPTER   X 
HIGHEST   COMMON   FACTOR 

142.  A  common  factor  of  two  or  more  algebraic  expressions 
is  an  expression  that  divides  each  of  them  without  a  re- 
mainder. 

Thus :  asbs,  a354,  and  a?b5  may  each  be  divided  by  ab. 
Therefore,  ab  is  a  common  factor  of  a363,  a364,  and  a2bb. 

In  this  definition  the  algebraic  expressions  are  understood 
to  include  only  rational  and  integral  expressions. 

143.  Expressions  having  no  common  factor  except  1  are  said 
to  be  prime  to  each  other. 

Thus  :  3  abx  and  7  cmn  are  prime  to  each  other. 


144.  The  highest  common  factor  of  two  or  more  algebraic  ex- 
pressions is  the  expression  of  highest  degree  that  divides  each 
of  them  without  a  remainder. 

Thus, 

{a363 
3  4  a263  is  the  expression  of  highest  degree  that  will 

...      divide  each  of  the  three  without  a  remainder. 
a2b5 

That  is,  a2b3  is  the  highest  common  factor  of  a3&3,  a364,  and  a?b&. 

145.  The  highest  common  factor  of  two  or  more  expressions  is 
the  product  of  the  lowest  powers  of  the  factors  common  to  the 
given  expressions. 

The  abbreviation  "  H.  C.  F."  is  commonly  used  in  practice, 

119 


120  HIGHEST  COMMON  FACTOR 

THE  H.  C.  F.  OF  MONOMIALS 

146.  The  H.  C.  F.  of  monomials  is  readily  found  by  inspection. 

Oral  Drill 
Give  orally  the  H.  C.  F.  of : 

1.  a364anda265.  8.    15  mnz  and  20  m2x. 

2.  m2nx2  and  mn2x.  9.    12  msn7  and  15  m4n9. 

3.  2m¥and4m¥.         10.   8  czdzm  and  12  <j*#n. 

4.  3  c3c?5  and  6  c2a?.  11.   35  arfy2z  and  42  m2?/2. 

5.  3  atmy3  and  9  am2?/2.     12.   51  asy2z3  and  17  a?/z2. 

6.  10  a2m2n2  and  15  m2n.   13.    12  arfyz,  16  afysz2,  and  20  aj?/22. 

7.  16  x*y  and  24  ?/22.  14.    18  cdm,  24  cmn,  30  cdn,  and  36  dmn. 

15.  5  c%,  10  cd32/,  15  cdy3,  and  20  c2dfy 

16.  33  mnx,  44  mw?/,  55  ma;?/,  and  66  mnxy. 

THE  H.C.F.  OF  POLYNOMIALS  BY  FACTORING 

147.  The  H.  C.  F.  of  factorable  polynomials  is  readily  found 
by  inspection  of  the  factors. 

Illustrations : 

1.  Find  the  H.  C.  F.  of 

a3 - 3a2 - 10 a,  a3 -  8 x2  + 15  a,  and  a3- 25a. 

Factoring,  a8  -  3  a2  -  10  a  =  a(a  -  5)  (a  +  2) 

a3  -  8  a2  +  15  a  =  a(a  -  3)  (a  -  5) 

a3-25a  =  a(a  +  5)(a-  5) 


Therefore,  H.  C.  F.  =  a{a  -  5)     Result. 

2.   Find  the  H.  C.  F.  of  m3  -  27,  9  -  m2,  and  m3  -  6  m2  +  9  m. 

Factoring,  wi8  -  27  =  O  -  3)  (m2  +  3  m  +  9) 

9-  m2  =  -(m  +  3)(m-  3) 
ro8  -  6  m2  +  9  m  =  m(m  -  3)2 


Therefore,  H.  C.  F.  =  (m  -  3)     Result. 

The  student  will  recall  that 

(9-m2)  =  (3  +  m)(3-ra)  =  (wi  +  3)(-m  +  3)=  -(ra  +  3)(m-3). 


H.C.F.   OF  POLYNOMIALS  BY  FACTORING  121 

Exercise  39 

By  factoring  obtain  the  H.  C.  F.  of  : 

1.  am  +  m,ax  +  x.  8.  a2  —  1,  (1  —  a)2. 

2.  am  +  m,  mx  +  m.  9.  a3  4- 1,  (1  -f  a)2. 

3.  cx  —  dx,cy  —  dy.  10.  c3  —  1,  2^-f  2c  +  2. 

4.  mx  +  m,mx  —  m.  11.  c3  —  cd2,  c3  —  2  c2^  +  cd?. 

5.  c*  +  c,  c2  —  c.  12.  4arJ  — 4  a,  aar*— 6a#4-9a. 

6.  m2  +  m,  ra2-l.  13.  c3-4c2-12c,  2  (^  +  4  c2. 

7.  0^-9,^  +  33;.  14.  27m-m4,  m3+3m2+9m. 

15.  ^  +  30  +  2,  c»-l,  c*  +  c. 

16.  §-x2yi,2xiy2-&xy,x2tf-xy-§. 

17.  7c?-Ucd  +  7d2,Uci-Ud?. 

18.  c6-l,  c3_c,  c4-l,  1-c3. 

19.  a4-«4,  a4  +  5aV  +  4a;4,  a4  +  aV. 

20.  m2  +  ma;  —  m  —  #,  m*  —  m,  m4  —  m3. 

21.  ,c2-3c  +  2,  c«_c-2,  6-c-c2. 

22.  2  m3  +  4  m2  -  30  m,  4  m3-  20  m2  +  24  m,  6  m3- 12  m2-18  m. 

23.  8  a3  +  64  b3,  12  a3  +  48  a2b  4-  48  a&2,  96  a2b2  -  24  a4. 

24.  m4  —  n4,  m3  4-  n3,  m5  4-  n5,  m3  4-  2  m2n  4-  mn2. 

25.  (^-(m  +  l)2,  (m4-c)2-l,m2-(c4-l)2. 

26.  4ar>-14z  +  12,  8^-32 a  +  30,  18 a2 - 69 x 4- 63. 

27.  a2  +  ax  —  2  a  —  2  #,  am  4-  a  4-  m#  4-  as,  a3  —  ax2. 

28.  6a2x  +  6a2-6x2-6x,4:a2m-4:a2-±mx+4:X,2a2x-2x2 
-2a2-\-2x. 

29.  3a2  +  6a2/4-3aa;4-6iB2/,  6a2  +  6ai/  +  6aa;  +  6a^,9a2-9a2/ 
4-  9  ax  —  9  #?/. 

130.   am  +  an  +  a  +  mx  +  nx  +  x,    m2  —  n2  +  m  —  n,    m2  +  m  4-  w 
4-  2  mn  4-  n2. 


CHAPTER  XI 
FRACTIONS.     TRANSFORMATIONS 

148.  An  algebraic  fraction  is  an  indicated  quotient  of  two 
algebraic  expressions. 

Thus  :    2,  ^,  a2  +  a6  +  62  are  algebraic  fractions. 
b       x  a-2b 

149.  The  numerator  of  a  fraction  is  the  dividend;  the  denomi- 
nator, the  divisor. 

The   numerator   and  the  denominator  are  the  terms   of   a 
fraction : 


The  following  principle  is  of  importance  in  processes  with 
fractions : 

Since,  by  definition,  a  fraction  is  an  indicated  quotient,  we 
may  let  the  quotient  of  a  divided  by  b  be  represented  by  x. 

Then 

b 
The  dividend  being  equal  to  the  product  of  the  divisor  by  the  quotient, 

a  =  bx. 

Multiplying  by  m,  am  =  bmx.  (Ax.  3) 

Considering  am  a  dividend,   bm  a  divisor,   and  x  a  corresponding 

quotient, 

a^  =  x. 
bm 

Hence,  ^m  =  a  (Ax.  4) 

bm      b 

150.  That  is  :  The  value  of  a  fraction  is  unchanged  when  both 
numerator  and  denominator  are  multiplied  or  divided  by  the  same 
quantity. 

122 


THE   SIGNS  OP  A  FRACTION  123 

THE  SIGNS  OF  A  FRACTION 

151.  Three  signs  are  considered  in  determining  the  quality 
of  a  fraction.     From  the  law  of  signs, 

a  _  —  a  _  _  —  a  _  _    a 
b~  -b~         b  -j>  ' 

Since  each  fraction  has  the  same  value,  -, 

o 

From  the  second  fraction : 

If  the  signs  of  both  numerator  and  denominator  of  a  fraction 
are  changed,  the  value  of  the  fraction  is  not  changed. 

From  the  third  and  fourth  fractions : 

If  the  sign  of  either  numerator  or  denominator  and  the  sign  of 
the  fraction  are  changed,  the  value  of  the  fraction  is  not  changed. 

152.  Consider  the  signs  of  the  factors  of  the  terms  of  a  fraction : 

■       (+<0       =|        (-a)  ■        (-a)  ■        (+o)  m 

(+&X+C)  (+&)(- <0  (-&)(+e)  (-&)(_<>) "       KJ 

Note  that  in  each  fraction  two  signs  are  changed  and  that  the  sign  of  the 
fraction  is  not  changed. 

+      (+«)       =  _      (-«)       -  _       (+«)       _.  _       (-«)  (2\ 

(+*)(+•)  C+^X+c)  (-6)(+c)  (-ft)C-c)'       W 

Note  that  in  these  fractions  one  sign  is  changed  or  three  signs  are 
changed,  and  that  the  sign  of  the  fraction  is  changed. 

(1)  We  may  change  the  signs  of  an  even  number  of  factors  in 
either  numerator  or  denominator  of  a  fraction  without  changing 
the  sign  before  the  fraction. 

(2)  We  may  change  the  signs  of  an  odd  number  of  factors  in 
either  numerator  or  denominator  of  a  fraction  if  we  change  the 
sign  before  the  fraction. 

The  following  is  a  common  application  of  these  principles : 

,  etc. 


(w  —  n)  (x  —  y)         (in  —  ri)  (x  —  y)      (n  —  m)  (x  —  y) 


124  FRACTIONS.    TRANSFORMATIONS 

TRANSFORMATIONS  OF  FRACTIONS 

To  reduce  a  Fraction  to  its  Lowest  Terms. 

153.    A  fraction  is  in  its  lowest  terms  when- the  numerator 
and  denominator  have  no  common  factor. 

Illustrations : 

!       45  aWc  _  3  x  3  x  5  aWc  =  3  ab2      Regult 
60  a2b2c     4x3x5  a2b2c        4 

2  a*  —  ax*  _  a(a  -  x)  (a2  +  ax  +  x2)  _  a2  +  ax  +  x2      Resuit 
a8  —  ax2  a(a  —  x)  (a  +  x)  a  +  x 

3  ab  +  bm  —  am  —  m2  _  (a  +  m)  (b  —  m)  _  —  (a  +  m)  (m  —  b) 

m2  -  b2  ~  (m  +  b)  (m  -  b)  ~     (m  +  b)  (m  -  b) 


a  +  m 


Result. 


m  +  b 
For  (b  -  m)=  (  -  m  +  6)  =  -  (w  -  6)  (Art.  152). 

In  general,  therefore,  to  reduce  a  fraction  to  an  equivalent 
fraction  in  its  lowest  terms : 

154.   Factor  both  numerator  and  denominator. 
Cancel  the  factors  common  to  both. 

To  cancel  is  to  divide  both  numerator  and  denominator  by  a  factor 
common  to  both.  The  expression  "cancel"  cannot  be  applied  to  any 
other  operation  in  algebra.  The  terms  of  an  expression  in  a  numerator 
cannot  be  canceled  with  like  terms  in  a  denominator,  for  such  an  operation 
is  not  division. 

Oral  Drill 

Eeduce  orally  to  lowest  terms  : 

8  m2  35afy  AZrfyz  48  c3fe2 

'  12m*  '  2Sxtf  '    28xyz'  '    72  c2d4x 

2    15  a&  39  m2n2  19  a2fzA  .       lUasx2z 

20ac*  '    65m3n*         '    76a3y2z'  '    95a3x'z' 

_     14m2yi  24  c3d2  q     69  m2ny  12    81  a2mn3x 

21  mn  '    36 cAd'  '  46 m2n2y2'  '    135  m2naT 


TRANSFORMATIONS  OF  FRACTIONS  125 

Exercise  40 

Eeduce  to  lowest  terms  : 

4a2-8a  10    m2-(x  +  l)\ 

4a2  +  20a*  '    (m-xf-1 

3m2-3n2  ■        (^-(c  +  l)2 

6m3-6r/  '    (c  +  l)2-^ 


1. 


3. 


2a*4-2a2  +  2a  9-(a-2)2 

a2  +  a  +  l      '  '   (3-a)2-4* 


4    8c3-8  13     ari  +  (a  +  c)a;  +  ac 

'4c2_4  as2  -|-  (-J7L  -f-  c)x  +  cm 

2a3-18a  4    s»-4a?  +  4-<? 

*   2(a-3)2'  '   ic2-4  +  4c-c2* 

a*-$a*  +  7x_  4^  +  4aa;  +  12a  +  12a; 

a?-x       '  '     6x2  +  6ax-9x-9a  ' 

2c3-16c2  +  30c  a3a;-3a2-aa  +  3 

2c3-llc2  +  5c*  '   9a?-aW-9  +  »"" 

4a3  +  4a2_24a  16m4-2m 

4a3-36a      '  '   32m5  +  8m3  +  2ra* 

4c3  +  5c2_21c  a2_2a5  +  62-a;2 

3c4  +  81c      '  '   «2-a2  +  &2  +  25a;' 


To  transform  a  Fraction  to  an  Integral  or  a  Mixed  Expression. 

155.  A  mixed  expression  is  an  expression  having  both  frac- 
tional and  integral  terms. 

Thus,  a  +  -,  x  —  1  H — ,  are  mixed  expressions. 

a  *  + 1 

The  principle  by  which  a  fraction  ivhose  numerator  is  of 
higher  degree  than  its  denominator  is  transformed  to  an  integral 
or  mixed  expression  depends  upon  Art.  75,  by  which 

ab  +  r  _ab  .  r _      ,r 
b      ~  6      6  b 


126  FRACTIONS.     TRANSFORMATIONS 

Illustration : 

£C3_|_'K2 o 

Change  — -^ — - —  to  an  integral  or  a  mixed  expression, 
ar  + 1 

x8  4-  £2  -  2(a2  +  1 
s3  +  x         («  +1 
+  x2  -  £C  -  2 
-fa2        +1 
-x~3 
Hence, 
xs  4  x*  -  2  =  (a*+ !)(«  +  !)- a; -3  =  (a;2  +  1)  (x  4  1)  ,  -s-3 
«2  +  l  x2  +  l  x2  +  l  x*  +  l 

By  changing  the  sign  of  the  numerator  the  form  of  the  result  becomes 

g  +  l  +  -»-g  =  a.4a  +  -(*  +  8):=s  +  l -■*+£.    Result. 
T  x2  +  l  x2  +  l  a2  4  1 

Therefore,  to  change  a  fraction  to  an  integral  or  a  mixed 
expression : 

156.  Divide  the  numerator  of  the  fraction  by  the  denominator. 
Write  the  remainder  over  the  denominator,  and  annex  the  result- 
ing fraction  to  the  integral  quotient  obtained.  If  the  sigyi  of  the 
first  term  of  the  remainder  is  negative,  change  the  signs  of  the 
entire  remainder,  and  the  sign  of  the  fraction  annexed  will  be 
negative. 

Exercise  41 

Change  to  integral  or  to  mixed  expressions  : 

c44-l 
1.     C-JLL±.  5. 

c-fl 

0     m2  +  m  + 1 :  6 


m  —  1 
a2-Sa-2 


7. 


a  —  5 

h4a- 
a2  +  a  —  1  m 


4     a4  -f-  4  a  +  1 . 


c3-c-3 

c2+2 

2n8-3n2  +  2 

n2  -  n ■  +  3 

a3  +  a2  +  a-7 

a2 +2 

3  m4  —  m2  —  ra 

+  1 

CHAPTER   XII 

FRACTIONS    (Continued)  — LOWEST   COMMON   MULTIPLE. 
LOWEST    COMMON    DENOMINATOR.     ADDITION 

THE  LOWEST  COMMON  MULTIPLE 

157.  A  common  multiple  of  two  or  more  algebraic  expres- 
sions is  an  expression  that  may  be  divided  by  each  of  them 
without  a  remainder. 

Thus,  a6&6  will  contain  a368,  a864,  and  a265. 

Therefore,  a666  is  a  common  multiple  of  a8&3,  a364,  and  a266. 

In  this  definition  the  algebraic  expressions  are  understood 
to  include  only  rational  and  integral  expressions. 

158.  The  lowest  common  multiple  of  two  or  more  algebraic 
expressions  is  the  expression  of  lowest  degree  and  least  numeri- 
cal coefficient  that  will  contain  each  of  them  without  a  re- 
mainder.    Thus : 

(3  asbs         30  a865  is  the  expression  of  lowest  degree  and  least 
2  a364      numerical  coefficient  that  will  contain  each  of  them 
5  a?b6      without  a  remainder. 
That  is,  30  a365  is  the  lowest  common  multiple  of  3  a863,  2  a8&4,  and 
5  a265. 

159.  The  lowest  common  multiple  of  two  or  more  expressions  is 
the  product  of  the  highest  poivers  of  all  the  factors  that  occur  in 
the  given  expressions. 

The  abbreviation  "L.  CM."  is  commonly  used  in  practice. 

127 


128  FRACTIONS 


THE  L.C.M.   OF  MONOMIALS 

160.  The  L.  C.  M.  of  monomials  is  readily  found  by  in- 
spection. 

Oral  Drill 

Give  orally  the  L.  C.  M.  of : 

1.  <?m  and  cm2.  7.  16  ah  and  24  a2c. 

2.  ra2?/3  and  m3y2.  8.  9  x2\f  and  18  a^i/V. 

3.  2  m2n  and  3  msn.  9.  10  c2dx  and  8  cd2y. 

4.  3  xy2  and  6  ax.  10.  16  mnsy  and  12  m2^2?/3. 

5.  8  my  and  12  wy.  11.  15  ah  and  20  cV. 

6.  10  ra2n2a;  and  6  mtfx2.  12.  25  aW  and  20  aWc5. 

THE  L.C.M.  OF  POLYNOMIALS  BY  FACTORING 

161.  The  L.  CM.  of    factorable   polynomials   is   readily 
found  by  inspection  of  the  factors. 

Illustrations : 

1.  Find  the  L.  C.  M.  of  a2  -  ab  and  ab  -  b2. 

Factoring,  a2  —  ab  —  a  (a  —  6) 

ab-b'2  =  b(a-b) 
Therefore,  L.  C.  M.  =  ab  (a-  b)    Result. 

2.  Find  the  L.C.M.  of  0?+ a  -  2,  a2- a- 6,  and  a?-4«  +  3. 

Factoring,  x2  +  x  -  2  =  (x  +  2)  (a  -  1 ) , 

SC*-x-6  =  (x-3)(x  +  2), 

x2  -  4  x  +  3  =  (x  -  3)  (x  -  1), 


Therefore,  L.  C.  M.  =  (x  +  2) (x  -  1) (x  -  3).    Result. 

3.   Find   the   L.C.M.   of    2  a3  +  8  a2  +  8  a,   ±x-a2x,  and 
5a2-20ci  +  20. 

Factoring,      2  a8  +  8  «2  +  8  a  =  2  a  (a  +  2)  (a  +  2) 
4  x  -  a2x  =  -  x  (a  +  2)(a  -  2) 
5  a2  -  20  a  +  20  =  5  (a  -  2)  (a  -  2) 
Therefore,  L.  C.  M.  =  10  ax  (a  +  2)2(a  -  2)2.     Result. 


THE  LOWEST  COMMON  DENOMINATOR  129 

Exercise  42 

By  factoring  obtain  the  L.  C.  M.  of : 

1.  x2  +  x,  xy -f  y.  ,        5.   2  a(a  +  1),  3a2  +  3  a. 

2.  am  —  a,m2  —  m.  6.   mV-9,m¥-6mn  +  9. 

3.  c*  +  c,c?-c.  7.   (W-l,  (W-l. 

4.  2/2  —  1,  (y  —  l)2.  8.   a3-4a2  +  4a,  a4-8a. 

9.  ^-ec5^^,  (^-81  c. 

10.  m3-f  3ra2  +  2m,  m3  +  4m2  +  3ra. 

11.  3  a;  (a?  +  a;  + 1),  4a4  +  4a. 

12.  7a3-175a,  3  a3  -30  a2  +  75a. 
•      13.  (x  +  y)2,  (y-xy,x>-tf. 

14.  tf(y-z\y(tf-#),y  +  z. 

15.  3(aj»  +  ajy),  8(^-  t/2),  12  (x2  -i/2). 

16.  2c3-c2-c,  2c3-3c2-2c. 

17.  5«4-5^,  4^-8^  +  40?. 

18.  7a3-7a,  4a(a-l)2,  3a3  +  6a2  +  3a. 

19.  m3  +  l,  3(2  +  3ra  +  m2),  4ra-4m2  +  4m8. 

20.  18-2^,3^-81,  4a;4-324. 

21.  am-f-a+  m  +  1,  am2  +  am  —  m2  —  m. 

22 .  5  c  (c  -  2) 2,  4  ca>  + 12  c  -  24  -  8  x,  24  c2  -  3  c5. 

23.  (a  +  xf-1,  a2-(x-l)2,  (a-lf-x2. 

24.  4a2-14a;  +  l2,  8  »* - 32  a  +  30, 18 as" - 69 »  +  63. 

25.  6  a2A5  +  6  a2  —  6  x2  —  6  a,  4  a2m  —  4  a2  —  4  m#  +  4  as, 

2a2<c-2z2-2a2'+2a. 

THE  LOWEST  COMMON  DENOMINATOR 

162.    The  lowest  common  denominator  of  two  or  more  algebraic 
fractions  is  the  lowest  common  multiple  of  their  denominators. 
The  abbreviation  "  L.  C.  D."  is  commonly  used  in  practice. 

SOM.    EL.   ALG.  — 9 


130  FRACTIONS 


Illustrations : 

1.    Change   — ,  — ,  and  -  to  equivalent  fractions  having  a 
o      4  o 

common  denominator. 

The  L.  C.  M.  of  the  denominators  3,  4,  and  6  is  12.     L.  C.  D.  s  12. 
Dividing  each  denominator  into  the  common  denominator,  we  have  : 

12  =  4,  «  =  8 ,!?  =  2. 

3  4         '6 

Multiplying  both  numerator  and  denominator  of  each  fraction  by  the 
respective  quotients  from  the  division,  we  have : 

5^  =  5^x4  =  20^  H 

3        3       4       12 
3z_3zx3  =  9z.    Regult 


4        4       3      12 

x_  x      2_2_a 
6       6       2      12 


x=x   x2  =  2_xi     Regult> 


2.    Change  — and  x„         to  equivalent  fractions  having 

af  +  x  xr  —  x 

a  common  denominator. 

The  L.  G.  M.    of   x2  +  x  and  x2  -  x    is    a:  (jc  +  1)  (x  —  1).      Hence 
L.  C.  D.  =  x  (x  +  1)  (x  -  1). 

Dividing  each  denominator  into  the  L.  C.  D.,  we  have  : 

x(x  +  l)(x~l)_x      1.  a  (g  + 1)  (a  -  1)  =  g  |  x 

X2  +  X  x2  -  x 

Multiplying  both  numerator  and  denominator  of  each  fraction  by  the 
corresponding  quotients  obtained : 

£rl  x  *=±=  I^zDL.    Result. 
X2  +  X      X-l      x(x'2-l) 

x  +  lxx+l=(x+l)2.     Regult 
x2-x     x  +  1     x(x2-l) 

In  general,  to  change  two  or  more  fractions  to  equivalent 
fractions  having  a  common  denominator: 


THE  LOWEST   COMMON   DENOMINATOR  131 

163.   If  necessary,  reduce  the  fractions  to  their  lowest  terms. 

Find  the  lowest  common  multiple  of  the  given  denominators, 
for  the  common  denominator. 

Divide  each  given  denominator  into  the  common  denominator. 

Multiply  both  numerator  and  denominator  of  each  given  fraction 
by  the  respective  quotients  obtained. 

Exercise  43 

Change  to  equivalent  fractions  having  a  common  denominator : 
3m   2m   5ra  2         3 

l"   T'    3  '    2   '  7' 

0     5a    3a    a 

2-  T'  T'  S'  8* 


AAA 

2a    4a'3a 


9. 


4.    A,  A,  1.  10 

ax1  bic9  ab 

5      ±       2-    ±  11 

'   a°b'  aV  aV 

6.    -i-.  -i-,  JU.  12 


c+1'  c- 
m 

1 

m 

m  +  n  n 
c 

—  m 

c 

<?-!'  (c 
2 

-i)8 

5 

a 

(y-1)2 
a 

3a  +  4' 

a; 

9  a2 -16 
a; 

m2n2/'  mny  '  mny2  '   oj8  —  1 '  a^  +  a^  +  o? 

3  2 


13. 


14. 


m2_m_e?  m2  +  2m-15 
1  2 


2c2-c-l'  6c2-c-2 
15 


16. 


17. 


a2  +  4a  +  3'  a2-a-12'  a 

1  2  3 

3+1'  (a  +  1)2'  (a;  +  l)3* 

c  c  3c 


3(c  + 1)'  4(c  -  1)'  2(c2  +  2  c  + 1) 


132  FRACTIONS 


18. 


19. 


20. 


m  m  m 


m2  +  m  m2  —  m  -j- 1'  m4  -f-  m 

3a;  4a;  5  a; 

2a7^2'  3a;  +  3'  4a^-4* 

1 1  1 

3^-15'  52/3-125y'72/  +  35' 


ADDITION  AND  SUBTRACTION  OF  FRACTIONS 

__         .    __  a  ,  b  .  c      a  +  b  +  c 

By  Art.  75,  -H h-=  — 

J  a;     a;     a?  a; 

That  is,  two  or  more  fractions  having  the  same  denominator 
may  be  added  by  adding  their  numerators,  and  writing  the  sum 
over  that  common  denominator. 

By  Art.  163,  any  two  or  more  fractions  may  be  changed  to 
equivalent  fractions  having  the  same  or  a  common  denominator. 

From  these  two  statements  it  is  clear  that  any  given  frac- 
tions may  be  added. 

The  term  "  simplify  "  is  used  to  include  both  operations  of 
addition  and  subtraction. 

Illustrations : 

x  x 


1.    Simplify 


a;  — 1     #  +  1 


The  L.  C.  D.  is  x*  -  1. 

Dividing  each  given  denominator  into  the  L.  C.  D.  and  multiplying  the 
corresponding  numerators  by  the  respective  quotients,  we  obtain  : 

x     _  (x  +  l)x  —  (x  —  l)x 


1      x  +  1  x*-l 

_  x*  +  x-x*  +  x 
x*-\ 

2x       _      u 
=  -2 — r     Result. 
x3  —  1 


ADDITION  AND  SUBTRACTION  OF  FRACTIONS       133 

o     _.      ...       2m3       9  ra  +  1 

2.    Simplify  —3 — r  — 2 '         . 

Every  integer  may  be  considered  as  having  a  denominator  1.  There- 
fore, in  adding  fractions  and  integers  the  integers  are  multiplied  by  the 
L.  C.  D. 

The  L. CD.  =  to3  -  1. 

Dividing  each  denominator  into  the  L.  C.  D.,  and  multiplying  as  be- 
fore, we  have : 

2  to3       2  to  +  1      _2  to3-2(to3- 1)  -  (to  +  1)  (to- 1) 

to3  —  1      1      to2  +  to  +  1  to3  —  1 

2  to3  -  (2  to3  -  2)  -  (m2  -  1) 

TO3—  1 

=l=J#.    Result. 

TO3  —  1 

In  general,  to  add  algebraic  fractions : 

164.    Reduce  the  given  fractions  to  lowest  terms. 

Divide  each  denominator  into  the  lowest  common  denominator, 
multiply  the  corresponding  numerators  by  the  quotients  obtained, 
and  write  the  sum  of  the  resulting  products  for  the  numerator  of 
the  result. 

The  sign  of  each  fraction  becomes  the  sign  of  its  numerator  in 
the  addition. 

Oral  Drill 


Simplify  orally : 

i.  M. 

X       X 

6. 

M-+2. 

XXX 

11. 

ax     ay     xy 

2.    ---• 
m     m 

7. 

a  ,b       c 
m     m     m 

12. 

a     b  ,  c 
x     y     z 

3    -4-^. 
3'   3  +  4 

8. 

3_2_a 

XXX 

13. 

2+1  +  -6. 
b            a 

4-  2+f. 

a     Z  a 

9. 

Sx     2x 

14. 

m     1      n 

n             m 

2  a     3  a 

x       2x 

10. 

J>_  _  3_ 

8  a     7 x 

15. 

i-^7 

134  FRACTIONS 


Exercise  44 

Simplify : 


i.  ~+x41-  i2.  -^-2  •   a 


a  +  1  a  — 1 

2     a  +  x~  3_l1~  a  +  x,  13  3 


4  a2  +  a  a2  —  a 

3  £±I_1_^+A  14         3c             4c 

2                    6  '    2c-2  3c  +  3* 

4  2m-l      m-1      7m+l  12               5 

5              3             10  "    7^-7     3a;2  +  3* 

5  c~ -      2      c  ~  ^-  16     m  +  3  m  + 1 

12                   6  '    m  +  4  m  +  2* 

a2  4- 2     a  — 1      a  +  l<  m  +  9  m  — 9 

4  a2   ""3a         2a*  *    m-9  m+9' 

a—b.b—c.c—a  ■  _     5  a  +  7  5  #  —  7 

7.    — r~  +  "T 1 18-   i s*—  •= =• 

a&         6c          ca  5<e  —  7  5sc+7 


x     x  +  y  c2  —  c  +  lc2+c  +  l 


9.    #-  +  -^«  20.      *«-*    4    2C  +  1 


2m     m  +  1  (2c  + 1)2      (2c  -  1) 

in       5             3                            01     m  — 5»  .      10m# 
io.    _ _ _.  4i,    ___  + — 


8xy     5xy—l  m  +  5x     m2  —  25  a2 

11.    X !£-.  22.  *"2  *+2 


ax     ax2-!  x2-2x  +  ±     a?  +  2x  +  4: 

1 


23. 


c^-f  7c2  +  12c     c8  +  8c2+15c 


24.    JL-h*-.     3 


ax  ax         2CUC2 


a  -J-  <c     a  —  a;     a*  —  x2 

26.    g3j|      3      *  +  l|(3*+7)tt 
x  +  1  *— 1        ar1  —  1 


27. 


28. 


29. 


ADDITION  AND   SUBTRACTION  OF  FRACTIONS        135 

a*  +  ra4  a4  m4 

"ay        aY  +  tf*     a4  +  ay* 
a-1        a  +  1       a2  4- 11 


2a  +  2     3a-3     6a2-6 

3a;  +  2        13a?-38        2a;-3 
a;  —  5       a2  — 4#  — 5       x  +  1 


30  4c2  2c         8(^  +  20  +  1 
*    4c2+2c  +  l"r2c-l  8c3-! 

31  a  — 1 a+1         | a-i 


32. 


a2-5a  +  6     a2-3o  +  2     a2-4a  +  3 
1  1  3a 


a^  +  2a;  +  4     ^-2^  +  4     a4 +  4^  + 16 


165.  The  process  of  addition  of  certain  forms  of  fractions 
is  simplified  by  application  of  the  principles  governing  the 
signs  of  fractions  (Art.  152). 

3  2  5a 


1.   Simplify 


a  -f-  1      1  —  a 


Changing  the  form  of  the  second  fraction,  in  order  that  the  terms  of 
the  factors  in  all  the  denominations  may  be  in  the  same  order,  we  have 
(Art.  152)  : 

2  2  2  2 


1  —  a          — a  +  1          —  (a  —  1)      a  —  1 

Therefore,      3            2           6a     _8(a- 1)  +  *«  +  lV-fi« 

a  +  la-1     «2_i                      a2_! 

—    ~~  '             ?           Rpfmlt 

a2  _  1        1  _  a2' 

2    Simplify             *                            *                            * 

1     "  (a-l)(a-c)      (l-a)(l-c)      (c-a)(c- 

-1) 

Changing  the  signs  of  both  factors  of  the  denominator  of  the  second 
fraction,  we  have  changed  the  signs  of  an  even  number  of  factors,  and 
the  sign  of  the  fraction  is  not  changed.  Changing  the  sign  of  one  factor 
in  the  denominator  of  the  third  fraction  changes  the  sign  of  the  fraction. 


136  FRACTIONS 

Therefore, 

1  1 


(a-l)(a-c)     (l-a)(l-c)     (c-a)(c-l) 
1  1 


(a  -1 )  (a  -  c)      (a  -  1)  (c  -  1)      (a  -  c)  (c  -  1)' 
L.C.D.  =(a-l)(a-c)(c-l) 

=  (c  -  l)  -  (g  -  C)  +  (a  -  1) 
(a-l)(a-c)(c-l) 

m 2c-2 

(«  -  1)  («  -  c)  (c  -  1) 

= ? .    Result. 

(a-l)(a-c) 

Exercise  45 
3c  2  2 


Simplify 


3. 


8. 


1_C2        c_l         c  +  l 

1  1_  10a? 

5x  +  l     5x-l     1-25  a^ 

m2  +  3m  +  9     m2-3m  +  9 
m  +  3  3  —  m 

4    c  —  2d  ,  2c  —  d  .  3(cy  —  ay) 

.■_  m  +  ft  .  a  +  ft 

o. —  ■ 

(m  —  a)(c  —  m)      (a  —  m)(c  —  a) 

i + i + i 

(a,  -l)(x-2)     (x-  2)  (3  -  a;)  T  (1  -  a?)  (a?  -  3) 

a;  +  a 2(c  —  a)  x  +  c 

(a?  —  a)  (#  —  6)      (c  —  a)  (a;  —  a)      (b  —  x)  (c  —  a;) 

1.1  1 


(b  —  c)(b  —  a)      (c  —  a)  (c  —  b)      (a  —  6)  (c  —  a) 

(1  —  a  —  a;)  a?(a;  —  1  —  a)  a(a  — 1  —  x) 

(x-l)(l-a)      (x-a)(x-T)      (a-l)(a-»)" 


CHAPTER   XIII 

FRACTIONS   (Continued)  —  MULTIPLICATION.     DIVISION. 
THE  COMPLEX  FORM 

MULTIPLICATION  OF  FRACTIONS 

(a)  A  Fraction  multiplied  by  a  Fraction 
The  product  of  two  fractions  is  obtained  as  follows : 
Given  two  fractions,  -  and  -,  and  let 


(3) 

(Ax.  3) 
(Ax.  3) 


X 

y 

m : 

=  2  (l)  andw=^(2). 
x                      y 

Multiplying  (1)  by  (2), 

mn  =  -  .  —  • 

x  y 

Multiplying  (1)  by  x, 
Multiplying  (2)  by  y, 
Therefore,  multiplying, 

mx  =  a. 
ny  =  b. 
mnxy  =  ab. 

Dividing  by  xy, 

ab 
mn  =  — 
xy 

But  from  (3), 

mn  =  - . — • 

x    y 

Therefore, 
In  general : 

a    b_abm 
x  '  y     xy 

(Ax.  4) 


(Ax.  5) 


166.  The  product  of  tivo  fractions  is  a  fraction  whose 
numerator  is  the  product  of  the  given  numerators,  and  whose 
denominator  is  the  product  of  the  given  denominators. 

(h)  A  Fraction  multiplied  by  an  Integer 
Since  any  integral  expression,  b,  has  a  denominator,  1 : 
From  Art.  166,  «x&  =  «.£  =  ^. 

X  X     1         X 

137 


138  FRACTIONS 

That  is : 

167.  A  fraction  is  multiplied  by  an  integral  expression  if  its 
numerator  is  multiplied  by  that  expression. 

The  process  of  multiplication  of  fractions  is  simplified  if 
factors  common  to  numerators  and  denominators  of  the  given 
fractions  are  canceled  before  multiplication. 

Illustrations : 

8c8c2    x  25 m2s4  =  8  ■  25  a*c*m*&  =  5a2c     Regult 


15  amsx8       16  ex       15-16  acmzx*       6  m 

The  cancellations  are :  8  in  16,  5  in  25  and  15,  a  in  a8,  c  in  c2,  ro2  in 
m8,  and  x*  in  x4. 

2.  Multiply  2+1*    ,("*-!)'     X5^i. 

m  —  1      m2  —  3m  +  2     m2  —  1 

Writing  each  fraction  with  numerators  and  denominators  factored, 

m  +  1  x      (m  -  l)2      x  m2-4  _  m  +  1  x  (m-l)Q-l)  x  (m+2)Q-2) 
m-1      m2-3m  +  2     m2-l     m-1      (m-l)(m-2)      (m+l)(m-l) 
Canceling  common  factors         =  m  +  2<  t       jjgsult. 


In  applying  cancellation  select  factors  for  divisors  from  the 
numerator  only.  Begin  at  the  left  of  the  numerator  and  seek 
a  possible  cancellation  for  each  new  factor  considered. 


Multiply  orally: 

Oral  Drill 

1.    aa%/x6m. 
3  m      ax 

5. 

17  c2      2n8 
6n4      51c4' 

2     Say  y  10c2 
5  c       9y 

6. 

5aa^      39^2 
13  xh     10  afy 

3     2  mn       9  c? 
3  cd      7  cm 

7. 

42^2x    10  z 
15  xy  '  28  afyz 

4     5m2x21c2d# 
3  cd     10  am 

8. 

32  x2yK  27  c4m2 
9  <?x       16  my 

MULTIPLICATION  OF  FRACTIONS  139 


cd 


Exercise  46 

Simplify : 

o  +  l     ^  (c  +  1?     (c-1)' 

*'    a-2     a2-l  c  c         (c2-!)5 

2.    ^X    /  +  1     ■  5.    ?^l.(?*-2x  +  ±). 
05  +  1      a^-^  +  a;  ar*  +  8    v  y 

0     m2— mn  ^  m2— 4mn+3  w2  fl     a^4-2a;  +  4     a^  +  8 


(m-rc)2  m2-n2  a?-4  ^-8 

7     3(c  +  l)2     7(c-l)3     4(c  +  l) 
'    2(c-l)2     6(c  +  l)2         7  c 


8. 


^-9a;4-18         ^-9 


^-27        ^-3aj-18 


•     c2-7c+12N/  c?-c-2 


c2_3c_4      ^-50+6 
10     3-4a+o2x        a3-16a 


5a  +  4     a3-7a2  +  12a 

n.      ^=9    x(a2  +  2a-15)xa2-5a  +  25 
a3  +  125      v  y  a  +  3 

m2  +  2  mn  +  n2  —  x*   m2  —  2  mx  +  x*  —  n2 


12. 


13. 


(m  -f-  n  -f-  a;)2         m2  —  2  mn  +  n2  —  x* 

c  +  3       c4-81    c*-9    (3-c)2 
(c-3)2'(c  +  3)2'c2  +  9    (c-3)2' 


14  ^—9      am +  2  a     am  —  2m  —  2a  +  4: 
m2-±      bx-3b       ax  +  3a-2x-6 

15  s3-8yS'-2ft  +  4     g  +  2 
*    a?  +  8     aj»  +  2a?  +  4     a>-2 


16. 


3tf2_a;_2'       3a4-3a;  10-10<e 


3*»+2*     5a^-10a;  +  5    6^  +  6^  +  60; 


140  FRACTIONS 

DIVISION  OF   FRACTIONS 

(a)  A  Fraction  divided  by  a  Fraction 

The  quotient  of  a  fraction  divided  by  a  fraction  is  obtained 
as  follows: 

We  will  assume  that  -  +-  -  =  m.  (1) 

x     y  w 

Now  a  dividend  equals  the  product  of  the  corresponding  divisor  and 
quotient. 

Therefore,  -  =  -  x  m. 

x     y 

Multiplying  by  £ ,  2  x  %*:$  x  m  x  *.. 

6  x     b     y  b 

Whence,  -xy=m.  (2) 

x     b  ,        v 

Hence,  from  (1)  and  (2) ,        5  -  5  =  5  x  y-  • 
w         v  "        x     y     x     b 

In  general : 

168.  The  quotient  of  a  fraction  divided  by  a  fraction  is  the 
product  of  the  dividend  by  the  inverted  divisor. 

169.  The  reciprocal  of  a  quantity  is  the  quotient  obtained  by 
dividing  1  by  that  quantity.     Thus  : 

-  is  the  reciprocal  of  a. 
a 

1  -f-  -  =  -,  the  reciprocal  of  -. 
3     2  3 

(b)  A  Fraction  divided  by  an  Integer 

From  Art.  168,  £  --  y  =  2  +  ^ 

a;  x     1 

a     1 

=  -  x  - 

x     y 

xy 


DIVISION  OF  FRACTIONS  141 

In  general : 

170.  The  quotient  of  a  fraction  by  an  integer  is  the  product  of 
the  given  fraction  by  the  reciprocal  of  the  integer. 

The  first  step  in  a  division  of  any  fraction  is,  therefore,  the 
inversion  of  the  divisor;  whence  the  process  becomes  a  multi- 
plication of  fractions. 

Illustrations  : 

1.  Divide  JJ&  by  M. 

S2cy    J    40  cy 

7  a2x  .  21  q2s  _  7  q2x      40  cy  _  5  x       Result 
32  cy  "    40  cy      32  cy      21  a2z     12  z 

2.  Divide   q2-a"2   by   a2~3a~4  . 

a2_5a+6    y  a2-8a  +  15 

q2_g_2    |   a2-3a-4  =  (a-2)(g  +  l)  .  (o-4)(g  +  l) 
a2  -  5  a  +  6  '  a"  -  8  a  +  15     (a  -  2)  (a  -  3)  '  (a  -  3)  (a  -  5) 
=  (q+l)x(a-3)(a-5) 
(a -3)      (a-4)(a+l) 
a  - 


a-4 


Result. 


3.    Divide    -   ^  +  27 by  ^-3^+9. 

*»  +  27       ^,(x2_3a;+9)=       x3+27  1 

x»  +  6x2  +  9x  x3  +  6x2  +  9x     aj2_3iB  +  9 

=  (s  +  3)(a«-3g+9)x  1 

a;  (x  +  3)  (a  +  3)         x2  -  3  x  +  9 

= Result. 

x  (x  +  3) 

Exercise  47 

Simplify : 

1  a2-^  .      a-x  (m2-!)2 
a2  +  ar>  '  3a*+3aj»"  m3 

2  c2-9  c-3  (ft  +  2)2  .     a? -4 
c2  +  2c  '  c2  +  3c  +  2*               '      a?-5     *  as8 -125' 


4-  (m2  - 1). 


142  FRACTIONS 


k     c2  — ccc— 6ar*  .  c+3x  x2  —  4a;  +  3  t  x  —  1 

(P-9CX2     '  c+2x  3-x        '  2  —  x 

n     a?  +  6x-7  .  x2-3x+2        0     aj"-l.      1-x 


x*-±x-2l      x*  +  x-6  1-x3     ar'  +  l+a; 

^_8aj  +  15     lx-10-x2 


9. 


10. 


6  —  50?  +  ^  '  x2  —  4a;+4 
(a.-l)2_a2  ^_(i_a)2 
(a._a)2_i  '  ^_(a_i)2' 


X1    a2-3ax<i2  +  3a  +  9  .    a3-27 


12 
13 


a2  +  3a  a+3  (a  +  3)2 

^-^-12     g^-2a;-3  .  a^+a; 
a^-9         a?2-2a;-8  '  x-2 

a?  —  ax  —  a-\-x_a?  —  8  8  —  a3 


2-3a  +  a2        a3-^     a?  +  ax  +  x* 

14    a2  +  lxa2  +  (a  +  l)2(a-l)2  .  ag  +  l 
'   a2-l  a2(a2 +1)+1        '  a3-l 

171.  When  addition  and  subtraction  with  multiplication  and 
division  occur  in  the  same  fractional  algebraic  expression  the 
additions  and  subtractions  are  first  performed. 

Illustration : 

Simplify  f^+l-^U^.-G  +  ^-Y 
*     J  \x-2     x  +  2J      V  x  +  2j 

(x  +  1     x-l\.(3x     6        12   \     r(x  +  l)(x  +  2)-(x-l)(x-2)-l 
\x-2     x  +  2/  '  V  £  +  2/     L  x2-4  J 

.  r(3s-6)(g  +  2)  +  12l 
L  x+2  J 

6  a;         3  x2 


a;2  -  4     x  +  2 
6s        a  +  2 
x2-4  '    3a:2 

: - Result. 

*(»-2) 


DIVISION  OF  FRACTIONS 


143 


Simplify : 


Exercise  48 


1. 


fx     a\f  aV  y 
\a     xj  \x  -f-  a  J 

V3      m]\m2 


3m 


6m  +  9 
6a; 


> 


'   \x        J\m2-2mx-3x2J\3        J 
\a  +  z       J\a  —  z        J     ar  —  z2 


(2.  +  «  +  D(8.-7  +  ?)  +  (.-g. 


+2     y\ 
(c  +  2y-(P    /         2d    Y 

(c  +  d  +  2f      \   ^c  +  2-d) 

10.  /r2a?  +  5  +  ?V3a-7  +  - 

11.  (c  +  2y-(c-2)';     8c  ' 


3m  +  3 


14.  r  *4  i^-iu^+i+^j. 

LC^-ox2)2]^6       )     W  a1) 

Vm+i^  J\m-i    A™-1    A^+i    / 

[2Va  +  2     a-2y/     2a2+4aJ     V  ; 


144  FRACTIONS 


[_mJ     n      mn       J     |__ftMr     mn     m'nj 

"  [(•^)("8(-+;)>'w+*w'+1' 


ti4 ! 

a-\-xj      c(acx  +  a  +  x) 

/'/7_1^2       ,v7_/*_9/7_i_9 


\       cy \acx -f- 

c4_2c3-c  +  2     (d-1)2  .  cd-c-2d  +  2 
±-d2  '    c2-c    '  d  +  1 


FRACTIONS  IN  THE  COMPLEX  FORM 


172.  A.  complex  fraction  is  a  fraction  whose  numerator  or 
denominator,  or  both,  are  fractions. 

The  order  of  processes  used  in  simplifying  complex  fractions 
varies  with  different  types,  and  no  general  statement  will  cover 
all  possible  cases  that  arise.  The  student  will  easily  under- 
stand the  following 

Illustrations : 

1.   Simplify      f     ■      x  * 


1_1  c3  +  d3 

d     c 

c2  +  cP  c2  +  d?  -  cd 

d  yc*-cP_          d          xc2-d* 

1_1  c8  +  #          c-d          c3  +  d3 

d     c  cd 


tf-cd+d*       cd     '        (c  +  d)(c-d)       _c 
d  c-d     (c  +  d)(c2-cd  +  d2) 

x 


2.    Simplify    *  +  2     x~2, 


x-2     x+2 


The  L.  C.  D.  of  both  numerator  and  denominator  is  (x  +  2)(x 
Multiplying  both  numerator  and  denominator  by  (x  +  2) (a;  —  2), 


FRACTIONS  IN  THE   COMPLEX  FORM  145 


x      .      x 
x  +  2     x-2  _x(x-2)  +  x(x  +  2)  _x2-2x  +  x2  +  2x_2x2  _x 
~lc        ~        x(x  +  2)  -x(x-2)     x2  +  2x-x2  +  2x      4x      2* 


x  -  2     x  +  2  Result. 

3.    Simplify  1  + ^ 

(The  work  begins  by  first  sim- 


■j    .       2         plifying  the  lowest  fraction,  etc.) 
1  —  x 

i+1   \  =i+t-V=i+7to=1+^ 

t  ,      2  3-s  3-x  3-x 

1-x  1-x 

=  1  +f_=^  =  _i_.     Result. 

1+x     l+x 

Exercise  49 

Simplify : 

6.    ^-1-     "-1     . 


1. 


•*■■■   M.  ■'           s 
x  —  o 

.     3        \ 
x  —  o 

x  +  3         2 

2         x  +  3 

1         1 

2     z  +  3 
c-1 

2o-c-±| 
c  +  1 

x*-tf-l      ^^ 

2a 

<C  +  1 


7. 


x     x+1 ,x— 1 

2     a-l  +  a;  +  l 

5-6s2  +  a4 

^Zd  +  l      £=2  +  1 

^+4_^^+2_ 

a^-4  *  „      x-2 


9    c2  +  4     c2 


ar*  +  4  x  +  2 

4      c2  +  4 


a*-ar*  +  l  c-2  c  +  2 

2a  c+2  c-2 

1-1  ,-1  2  +  1  x-1- 

m2-9  3     m  2  y j 


m3  +  27      m2  +  9  0     1  ,1  1 

-—Z m  2--  2/  +  -  y-- 

3                                        a?  2  a; 

SOM.    EL.    ALG.  — 10 


146 


FRACTIONS 


11. 


12. 


m 


m 


+i 


i+» 


l+m  +  m! 

i  +  2 


■i-(H-m)2. 


c  — 


2  + 


13. 


14. 


a  +  1 


—  1 


a-1 


+  1 


a-1 
2c  +  <b 


Hi 


a  +  1 
1 


-1 


a-1 


a  +  1 


2c  +  x 


2  ex 


2c  +  x 


U-x2 


15. 


16. 


17. 


18. 


a  +  2       2       1      a(2-a) 
l-2a  l+2a 


l  +  2("  +  2)       a  +  !^L 
T  l-2a  ^l  +  2a 


1  + 


a  —  1     a+1      a— 1 


a  +  1 
a-1 


a—1      a+1      a—1      a+1 

a+1 . a—1 


-1  + 


a+1      a—1      a+1 

1  +  A+I 

9      3a      a2 


[~a3  +  27  .  a2  +  9~|  .  f(a-3)2  +  3a  .  ft      IV] 
[a8-27  ■  a2-9j  '  [_(a+3)2-3a  '  \9     a7J 

U  +  2        JL       (*  +  2)8J 


(i rnL\  (  ^    +i+^_A 

V       (x  +  2yj  V(*  +  2)2        ^x  +  2j 


CHAPTER  XIV 

FRACTIONAL  AND  LITERAL  LINEAR  EQUATIONS 
PROBLEMS 

173.  To  clear  an  equation  of  fractions  is  to  change  its  form 
so  that  the  fractions  shall  disappear.  This  change  is  ac- 
complished by  the  use  of  the  L.  C.  D.  of  the  given  fractions. 

Given;  «±«=*±«>. 

4  o 

Multiplying  both  members  by  12, 

12  (x  +  6)      12  (x  +  10)t 

4  6 

Reducing,  3  (*  +  6)  =  2  (x  +  10). 

And  the  original  equation  is  merely  changed  in  form  and  is 
free  from  fractions. 

In  general,  to  clear  an  equation  of  fractions : 

174.  Multiply  both  members  of  the  equation  by  the  L.  C.  D.  of 
the  fractions,  remembering  that  the  sign  of  each  fraction  becomes 
the  sign  of  its  numerator.     Solve  the  resulting  integral  equation. 

Illustrations :  # 

I.   Solve^±l_^z2  =  6-£. 
3         '     5  2 

The  L.  C.  D.  =  30.     Multiplying  both  members  by  30, 
10(2x  +  l)-6(3x-2)=15(6-x). 
20  x  +  10  -  18 x  +  12  =  90  -  15x. 

20x  -  18x  +  15x  =  -  10  -  12  +  90. 
17x  =  68. 

x  =  4.     Result. 
147 


148  FRACTIONAL  AND  LITERAL   LINEAR   EQUATIONS 

x-1      x  +  2      „        2x? 


2.    Solve 


2- 


x  +  l      x  —  1  x2  —  l 

Multiplying  both  members  by  the  L.  C.  D.,  (%  +  l)(x  -  1), 
!)•-(*  +!)(*  +  2)  =  2(x2  -  1) 


x2-2a;  +  l 


From  which 


x2-Sz-2  =  2x2 
1 


2  a*. 

-2x2. 


x  =  —    Result. 
5 


Solve : 

x  4-1  ,a5  —  l 

3 
*-l 


1. 


3. 


4. 


4- 


4 
*  +  l 


2 

a; +  1 
3 

x-1 


=  4. 


3. 


Exercise   50 

'     2x-l      A-x     2x+l 
o. 

6. 


3 
2*4-1 


5 
3*4-1 


14 

15' 


7. 


=  4:'       8. 


3 

5            10 

4 

»-l 

Q      2*4-1 

3 

J_      5 

3 

x  +  2 

3*4-1     2*4-1 

4 

3               2 

3 

a;  4-2 

*-6     1-2* 

=  0. 


4-1. 


9.  i.(*_l)_f(*4-l)4-2=0. 

10.  f(2*-l)-TV  =  -f(5  +  3*). 

11.  (*4-|)(^-i)  =  *2. 

12.  (x-l)(x  +  i)  =  (x-l)(x-i). 

13.  i(*_2)-|(a;-r-l)-i(*-2)  =  0. 


14.    2f 


*-fl 


2x- 


15. 


(*4-l)2'    (*4-2)2^14-*! 


rA 


16. 


17. 


18. 


3 

x  —  6 


#4-5     *+3 
2*-3     4*-5 


3*  +  4 
3*-2 


6*-7 
2*-5 


6*  —  1      4*4-1 


=  0. 


19. 


20. 


21. 


4*-l      3*-l 

a?  4-1  "  x-1 
4  *  * 


1. 


*  — 5     x 
5*-l 


3  =  0. 


*4-3 


4-2 


7*-3 

*4-l 


=  0. 


SPECIAL  FORMS  OF  FRACTIONAL  LINEAR  EQUATIONS      149 
00       2x         Sx  x       .      2x  5 

Z6. — -J-  — — —  -+- 


23. 


»  +  l     a?  +  l     2a;  +  2      3a?  +  3  6 

a^  +  2a?  +  4        2  a;    ^-2^  +  4 
a  +  2  4-a2  x-2 


•24.    _6 2      _      6. 


25. 
26. 

27. 


2a;  +  3     3-2a     4a2-9 

3a;-4      18g2  +  a?  =  3a?  +  4 
3a>  +  4      9^-16     4-3z* 

9a^  +  3a?  +  l      6a;==9a;2-3a?  +  1 
3a;  +  l  l-3a; 

2  3  4 


<e2  +  3»  +  2     ^  +  405  +  3     x*  +  5x  +  6 

28.         *  +  1       + £-1 =       *  +  *       , 

2^-a;-l      2x2-3x-2     ^-3^  +  2 


SPECIAL  FORMS  OF  FRACTIONAL  LINEAR  EQUATIONS 

(a)   The  Independent  Monomial  Denominator 

Illustration : 

175.   Solve   ^±i_A^l_^-l  =  7. 
3  2a  +  l  2  6 

The  L.  CD.  of  the  monomial  denominators,  3,  2,  and  6,  is  6.  Multi- 
plying both  members  of  the  equation  by  6,  we  have, 

2(3x  +  4)  _6(a:-1)-3(2a;-l)  =  7. 

From  which  6  x  +  8  -  6  (z  ~  *)  -  6  x  +  3  =  7  • 

2x+  1 

At  this  point  note  that  all  x-terms  outside  of  the  fraction  disappear. 

In  any  simple  equation  of  this  type  the  unknown  term 
similarly  disappears  when  the  equation  is  cleared  of  monomial 
denominators.  If  the  unknown  term  does  not  disappear  except- 
ing from  the  fraction  having  the  binomial  denominator,  error 
has  been  made  in  the  work. 


150        FRACTIONAL  AND   LITERAL   LINEAR   EQUATIONS 
Transposing  and  collecting, 


Dividing  by  —  2, 
Clearing  of  fractions, 


6(s-l)=      j 
2x  +  l 


2x  +  l 

3(x-l)  =2(2x  +  l). 
3x-3  =  4x4-2. 

x  =  —  5.    Result. 


Solve 


1. 


2. 


4. 


5. 


6. 


7. 


8. 


9. 


Exercise  51 

x  4-1      a;  —  1  _  x—  1 
3         «4-l"~      3 

3a?-l      a?  +  l==2a?  +  lt 

3  a;  4- 2  2 

4a?+5     x  +  l  =  2x  —  l 

4  »— 1~      2 

2fl  +  l       a;—  1      5fl?  +  l=Q 

2  z  +  10  5 

2x  —  l      ^z2l.  —  alZll  —0 

4  2^4-4         2     ~~   ' 

3a;  — 4      9a;  — 4        a;  4- 11        x 

3  12  12a; -27     4 

2a?4-7  ,    x      Sx  —  1      5a;4-6 


=  0. 


12 


3a;-3 


10. 


3a?4-5  4a?-l      2a?  +  l  =  75 
2a;-5  6  3  6' 

8a;-5  2j     16  a;  +  3  =  x-H 

8  2  16  a?-5* 

2  a;- 5  3a?  +  l  =  3a?  +  £,3a?  — 7 
4  8  a?— $■  24 


SPECIAL  FORMS  OF  FRACTIONAL  LINEAR  EQUATIONS      151 

(b)  Forms  having  Four  or  More  Dissimilar   Denominators 

176.    Illustration : 

0 ,      x— 3     x—2     x—7      x  —  6 

Solve -  = -• 

x—2      x—1      x—b      x—o 

To  avoid  the  use  of  a  common  denominator  having  four  binomial 
factors,  each  member  of  the  equation  is  first  simplified  independently  of 
the  other. 

Combining  separately, 

(x  -  1)  (x  -  3)  -  (x  -  2)  (x  -  2)  =  (x  -  7)(x  -  5)  -  (x  -  6)  (x  -  6) 
(x  -  1)  {x  -  2)  (x  -  6)(x  -  5) 

imp    ymg,       _______  =  ______ . 

The  fractions  being  equal  and  having  the  same  numerators,  it  follows 
that  the  denominators  must  be  equal.    Therefore, 

(x  -  1)  (x  -  2)  =  (x  -  6)  (x  -  5). 
x2  -  3  x  +  2  =  x2  -  11  x  +  30. 
8  x  =  28. 
x=\.    Result. 


Solve 


3. 


Exercise  52 

1 

1 

1 

1 

„-5 

x  —  4 

x  —  3 

x-2 

1 

1 

1 

1 

x  +  1 

x  +  2 

x  +  3 

«  +  4 

2 

1 

x  —  4 

1 
x-1 

2 

2a-3 

2x- 

7 

11 

11 

7 

7 

6. 


x  +  3  x  —  4     a-j-2  x-9 

x  +  2  a;+3_a;  +  4  a;  +  5< 

a;+5  x+6~~x  +  7  x  +  S' 

x  —  S  x—  7  _x—  5  a;— 4 

#-10  a;-9~a;-7  #-6 


152     FRACTIONAL   AND  LITERAL   LINEAR   EQUATIONS 

(c)  Literal  Fractional  Equations 

177.  In  literal  equations  either  a  portion  or  all  of  the  assumed 
known  quantities  are  represented  by  letters,  the  first  letters  of 
the  alphabet  being  ordinarily  chosen  for  these  known  quanti- 
ties. 

Illustration : 

Solve£±^+i=l^. 

x  —  a  2x  —  a 

Multiplying  both  members  of  the  equation  by  L.  C.  D.  (x  —  a) (2 x  —  a), 
we  have, 

(x  +  a) (2  x  —  a)  +  (x  —  a) (2 x  -  a)  =  (4 x  -  a) {x  -  a). 

2x*  +  ax  -  a2  +2x2  -  3  ax  +  a2  =  4x2  -  5  ax  +  a2. 
2x2  +  2s2  -  4&2  +  ax  -  3  ax  +  6  ax  =  a2  -  a2  +  a2. 

3  ax  =  a2. 


Result. 


Exercise  53 


Solve : 


1.  3  a  — 4:  ex  =  7  a  — 6  ex. 

2.  (c  +  d)»+  (c-d)a  =  4c2. 

3.  m(sc  +  n)  -J-  n(sc  +  m)  =  1  +  2  mn. 

4.  (a  +  a)(a  +  l)  =  (a;  +  a  +  l)2. 
5".  (a-3a-c)2  =  (a;  +  a  +  3c)2. 

6.  (m  —  no?) (71  —  m#)  =  ran(sc*  —  1). 

m     sc     a;     n 

~    cd  ,         1  .    , 

8. l.  a  =  -  -f-  dm. 

x  x 

xxx 

9. =ra— n-}-». 

mn     mp     np 

10.   — 7—  =  — f— 

x  +  n   '  x  +  m 


SPECIAL  FORMS  OF  FRACTIONAL  LINEAR  EQUATIONS      153 

2a  +  5c_5a;  +  4c 
2x  —  5c     5  a?  —  4  c* 

i2.  ra^n=2-Sx-p. 

n-\-x  x+p 

j£±%      1 _2_ 

10    26     Sx     26      3x 
13. 


14. 
15. 
16. 
17. 
18. 
19. 
20. 


1 

36 

3 
2x 

36      2x 

X 

c 

X 

a  +  1 

1- 

-a     a2-l" 

x  +  c 

_x  + 

c  +  2 

x  —  c 

X  — 

c-2' 

2x  — 
x  —  c 

i  i 

x+c 

x  —  a 

+  2_ 
-2 

x  —  a-\- 1 

x—  c 

X—  c  +  1" 

c 

d 

__c  —  d 

x—c 

X  — 

d        x 

ex 

-<*+*+i§i 

c-\-x 

1  n. 

x    y 

\     f             X 

21. 
22. 
23. 
24. 


d  — 2 ' x+l     d+2 ' 


x- 


x  —  1       2(1  — n)  _x  —  n 
x—n     n+l—x     x—X 

x  —  m      x  +  2m_      5mx  +  9m2 
x  —  3m     a  — 4ra     x2— 7mx+12m* 

1  1      =      1 1_      . 

x  —  2c     x  —  6c     x  —  4:C     x  —  Sc 


154     FRACTIONAL  AND   LITERAL  LINEAR   EQUATIONS 

PROBLEMS  LEADING  TO  FRACTIONAL  LINEAR  EQUATIONS 

Exercise  54 

1.  The  difference  between  the  fourth  and  the  ninth  parts 
of  a  certain  number  is  2  more  than  one  twelfth  of  the  number. 
Find  the  number. 

Let  x  =  the  required  number. 

Then  ?_^  =  the     difference    between    the 

fourth  and  ninth  parts. 


-*+2: 

12 

=  one    twelfth    the 

number   in- 

Hence,  from  the  conditions, 

creased  by  2. 

X     xm 

4     9' 

=  —  +  2. 
12 

Clearing, 

9 

X  —  4x  : 
2X: 

=  3a  +  72. 

=  72 

Verification 

:  4 

9" 

_36 

"  4  " 

36 
"  9 

X    : 
=  9-4: 

=  36,  the  required  number. 
=  5. 

X 

12 

+  2: 

_  36 
12 

+  2 

=  3  +  2: 

=  5. 

2.  Find  that  number  the  difference  of  whose  fifth  and  sixth 
parts  is  3  less  than  the  difference  of  its  third  and  fourth  parts. 

3.  When  the  sum  of  the  third,  eighth,  and  twelfth  parts  of 
a  number  is  divided  by  2  the  quotient  is  1  more  than  one  fourth 
the  number.     Find  the  number. 

4.  Find  three  consecutive  numbers  such  that  the  first 
divided  by  6,  the  second  by  5,  and  the  third  by  2,  give  quo- 
tients whose  sum  is  4  less  than  the  greatest  number. 

5.  The  first  digit  of  a  number  of  three  figures  is  two  thirds 
the  second  digit  and  4  less  than  the  third.  If  the  sum  of  the 
digits  is  18,  what  is  the  number? 

6.  The  sum  of  two  numbers  is  38,  and  if  the  greater  number 
is  divided  by  the  less  increased  by  2,  the  quotient  is  3  and  the 
remainder  4.     Find  the  number. 


PROBLEMS  WITH  FRACTIONAL  LINEAR   EQUATIONS      155 

Let  *  =  the  smaller  number. 

Then  38  —  x  =  the  greater  number. 

Now  Dividend  -  Remainder  =  Quotient< 

Divisor 

Hence,  <«-«)-*  =  8. 

'  x  +  2 

Or,  38  -  x  -  4  =  3  x  +  6. 

From  which  x  =  7,  the  smaller  number; 

and  38  —  x  =  31,  the  larger  number. 

Verifying:  (38  ^ ^ -  4  =  38  -  7  -  4  =  27  =  3 

3    *  x+2  7+2  9 

7.  The  sum  of  two  numbers  is  57,  and  if  the  greater  num- 
ber is  divided  by  the  less,  the  quotient  is  3  and  the  remainder  9. 
Find  the  numbers. 

8.  The  difference  between  two  numbers  is  23,  and  if  the 
greater  is  divided  by  4  less  than  twice  the  smaller,  the  quotient 
is  3.     Find  the  numbers. 

9.  The  sum  of  the  third,  fifth,  and  sixth  parts  of  a  number 
is  divided  by  one  half  the  number,  and  the  quotient  is  1  and 
the  remainder  6.     Find  the  numbers. 

10.  If  the  sum  of  three  consecutive  numbers  is  divided  by 
the  smallest  number  increased  by  7,  the  quotient  is  2  and  the 
remainder  is  6.     Find  the  three  numbers. 

11.  In  3  years  a  certain  man  will  be  half  as  old  as  his 
brother,  and  the  sum  of  their  present  ages  is  69  years.  What 
is  the  present  age  of  each  ? 

Let  x  =  the  man's  age  in  years  at  the  present  time. 

Then  69  —  x  =  the  brother's  age  in  years  at  the  present  time. 
Hence,  x  +  3  =  the  man's  age  after  3  years, 

and  72  —  x  =  the  brother's  age  after  3  years. 


Then 

3  +  3: 

72- a; 
2 

and 

69-z 

x- 

=  69  -  22  = 

=  22,  the  man's  present  age. 
=  47,  the  brother's  present  age. 

Verification : 

2(22  +  3)  = 
60  = 

=  47  +  3. 
=  50. 

156    FRACTIONAL  AND  LITERAL  LINEAR  EQUATIONS 

12.  A  is  one  third  as  old  as  B,  but  in  8  years  he  will  be 
only  one  half  as  old.     Find  the  present  age  of  each. 

13.  A's  age  is  one  fourth  that  of  B,  but  in  5  years  A  will  be 
one  third  as  old  as  B.     Find  the  present  age  of  each. 

14.  A  child  is  1|  times  as  old  as  his  brother,  but  2  years 
ago  he  was  1  \  times  as  old.     How  old  is  each  now  ? 

15.  The  sum  of  the  ages  of  a  father  and  son  is  72  years,  and 
if  the  son  were  one  year  older  and  the  father  one  year  younger, 
the  son's  age  would  be  one  third  that  of  the  father.  Find  the 
present  age  of  each. 

16.  If  the  length  and  the  width  of  a  certain  rectangular  field 
were  each  increased  by  10  feet,  the  area  of  the  field  would  be 
increased  by  800  square  feet.  If  the  length  is  now  10  feet 
more  than  the  width,  what  are  the  dimensions  of  the  field  ? 

Let  x  =  the  width  of  the  field  in  feet. 

Then  x  +  10  =  the  length  of  the  field  in  feet. 

Since  the  area  of  the  field  equals  the  product  of  the  length  by  the  width, 
x  (x  +  10)  =  x2  +  10  x  =  the  present  area  of  the  field  in 
square  feet. 
In  like  manner, 

(x  +  10)  (x  +  20)  =  x2  +  30  x  +  200  =  the  area  of  the  field  in  square 

feet  if  length  and  width  are 
increased. 
Hence,  (x2  +  30  x  +  200)  -  (x2  +  10  x)  =  800. 
x2  +  30  x  +  200  -  x2  -  10  x  =  800. 
20  x  =  600. 

x  =  30,  the  present  width  of  the  field ; 
and  x  +  10  =  40,  the  present  length  of  the  field. 

Verification :  (x  +  10)  (x  +  20)  -  x  (z  +  10)  =  (40)  (50)  -  30  (40) 

=  2000  -  1200 
=  800. 

17.  The  length  of  a  certain  rectangle  is  10  feet  greater,  and 
the  width  5  feet  less,  than  a  side  of  an  equivalent  square.  Find 
the  dimensions  of  the  rectangle. 


PROBLEMS   WITH  FRACTIONAL  LINEAR   EQUATIONS    157 

18.  A  square  has  the  same  area  as  a  certain  rectangle  whose 
length  is  20  feet  more,  and  whose  width  is  12  feet  less,  than 
the  side  of  this  particular  square.  Find  the  dimensions  of  the 
rectangle. 

19.  The  length  of  a  certain  rectangle  is  12  feet  greater  than 
the  width.  If  each  dimension  is  increased  by  3  feet,  the  area 
of  the  rectangle  will  be  greater  by  225  square  feet.  Find  the 
dimensions  of  the  rectangle  at  present. 

20.  A  square  is  cut  from  a  certain  rectangular  field,  the  length 
and  width  of  the  field  being  respectively  50  feet  and  30  feet 
greater  than  the  side  of  the  square  cut  out.  If  9500  square 
feet  remain  in  the  field,  what  was  the  original  length?  the 
original  width  ?  the  original  area  ? 

21.  A  can  do  a  piece  of  work  in  5  days,  B  the  same  work  in 
6  days,  and  C  the  same  in  7  days.  How  many  days  will  be  re- 
quired for  the  work  if  all  three  work  together  ? 

Let  x  =  the  number  of  days  all  three  together  require. 

\  —  the  portion  done  by  A  in  1  day. 

\  =  the  portion  done  by  B  alone  in  1  day. 

\  =  the  portion  done  by  C  alone  in  1  day. 
Hence,  \  +  \  +  \  =  the  amount  all  together  can  do  in  1  day. 

Therefore,  1  +  1  +  1  =  1. 

5     6     7     a; 

42s  +  35z  +  30x  =  210. 

107  x  =  210. 

x  s  Ifljf  days.    Result. 

22.  A  can  do  a  piece  of  work  in  4  days,  B  in  5  days,  and  C 
in  8  days.  How  many  days  will  be  required  if  all  three  work 
together  ? 

23.  A  and  B  can  together  build  a  wall  in  7  days,  and  C 
alone  can  build  the  wall  in  15  days.  How  many  days  will  be 
required  if  all  three  work  together  ? 


158        FRACTIONAL  AND  LITERAL  LINEAR  EQUATIONS 

24.  A  and  B  can  paint  a  house  in  8  days,  A  and  C  to- 
gether in  9  days,  and  A  alone  in  12  days.  In  how  many  days 
can  B  and  C  together  do  the  work  ? 

25.  Two  pipes  enter  a  tank,  the  first  of  which  can  fill  it  in 
7  hours  while  the  second  pipe  requires  9  hours  to  fill.  How 
many  hours  will  be  required  to  fill  the  tank  if  each  runs  alone 
1  hour,  and  then  both  run  together  until  filled  ? 

26.  At  what  time  between  4  and  5  o'clock  are  the  hands  of 
a  clock  together  ? 


At  4  o'clock  the  hour-hand  is  20  minute-spaces  ahead  of  the  minute- 
hand.    Hence, 

Let         x  =  the  number  of  spaces  the  minute-hand  passes  over, 
and     x  —  20  =  the  number  of  spaces  the  hour-hand  passes  over. 
Now  the  minute-hand  moves  12  times  as  fast  as  the  hour-hand. 
Therefore, 

12  (x  —  20)  =  the  number  of  spaces  the  minute-hand  passes  over. 
Hence,  12  (x  -  20)  =  x. 

12  x  -  240  =  x. 

\\x=  240. 
x  =  21  A- 
That  is,  the  hands  of  the  clock  will  be  together  at  21T9T  minutes  after 
4  o'clock. 

27.  At  what  time  between  3  and  4  o'clock  will  the  hands 
of  a  clock  be  together  ? 

28.  At  what  time  between  7  and  8  o'clock  will  the  hands  of 
a  clock  be  together  ? 

29.  At  what  time  between  10  and  11  o'clock  will  the  hands 
of  a  clock  be  at  right  angles  to  each  other?  (Hint:  In  this 
case  the  position  of  the  minute-hand  will  be  15  minute-spaces 
behind  the  hour-hand.) 

30.  Find  the  time  between  4  and  5  o'clock  when  the  hands 
of  a  clock  are  at  right  angles  to  each  other.  (Hint  :  Two  pos- 
sible positions  may  be  found  in  this  case,  for  the  minute-hand 
may  be  15  minute-spaces  ahead  or  behind  the  hour-hand.) 


PROBLEMS  WITH  FRACTIONAL  LINEAR  EQUATIONS    159 

31.  The  denominator  of  a  certain  fraction  is  greater  by  2 
than  the  numerator.  If  1  is  added  to  both  the  numerator 
and  denominator,  the  fraction  becomes  -|.  Find  the  original 
fraction. 

32.  Out  of  a  certain  sum  a  man  paid  a  bill  of  $30,  loaned 
I  of  the  remainder,  and  finally  had  left  $  56.  How  much  had 
he  at  first  ? 

33.  The  largest  of  three  consecutive  odd  numbers  is  divided 
into  the  sum  of  the  other  two,  the  quotient  being  1  and  the 
remainder  9.     Find  the  numbers. 

34.  The  sum  of  the  ages  of  a  father  and  son  is  80  years, 
but  if  each  were  2  years  older  the  son's  age  would  be  f  the 
father's  age.     How  old  is  each  ? 

35.  A  certain  number  is  decreased  by  12  and  the  remainder 
is  divided  by  4.  If  the  resulting  quotient  is  increased  by  7, 
the  sum  is  the  same  as  if  the  original  number  had  been  in- 
creased by  7  and  then  divided  by  3.     Find  the  number. 

36.  A  man  gave  -|  of  a  certain  sum  to  relatives,  y1^  to  each 
of  two  churches,  ^  to  a  library,  and  the  remainder,  $  60.00,  to 
a  hospital.     What  was  the  total  bequeathed  ? 

37.  In  a  certain  baseball  game  a  total  of  13  runs  was  made 
by  both  teams.  If  the  winning  team  had  made  2  more  runs, 
and  the  losing  team  3  less,  the  quotient  obtained  by  dividing 
the  winning  runs  by  the  losing  runs  would  have  been  5.  How 
many  runs  did  each  team  make  ? 

38.  The  distance  around  a  rectangular  field  is  96  rods,  and 
the  length  of  the  field  is  f  the  width.  Find  the  length  and 
the  width  of  the  rectangle,  and  the  number  of  square  feet  it 
contains. 

39.  In  eight  games  a  certain  fielder  made  2  less  runs  than 
hits.  If  5  times  the  number  of  hits  he  made  is  divided  by  the 
number  of  runs  increased  by  3,  the  quotient  is  4.  How 
many  hits  and  how  many  runs  did  he  make? 


160    FRACTIONAL  LITERAL  AND  LINEAR  EQUATIONS 


40.  Find  three  consecutive  even  numbers  such  that  3  less 
than  one  half  the  first,  plus  2  less  than  one  half  the  second, 
plus  1  less  than  one  half  the  third  equals  15. 

41.  In  going  a  certain  distance  an  automobile  moving  20 
miles  an  hour  required  3  hours  less  time  than  a  second  auto- 
mobile making  16  miles  an  hour.  What  was  the  distance  in 
miles  ? 

42.  A  number  is  composed  of  two  digits  whose  difference 
is  4.     If  the  digits  are  reversed,  the  resulting  number  is  ^  the  I 
original  number.     Find  the  number. 

43.  A  can  run  10  yards  in  1  second,  B  8  yards  in  1  second.  . 
If  A  gives  B  a  start  of  3  seconds,  in  how  many  seconds  will 
A  overtake  B  ? 

44.  The  length  of  a  rectangle  is  9  rods  more  than  its  width. 
If  the  length  is  increased  by  6  rods  and  the  width  decreased 
by  3  rods,  the  area  is  unchanged.  Find  the  length  and  breadth 
of  the  rectangle. 

45.  An  automobile  going  25  miles  an  hour  is  40  minutes 
ahead  of  one  going  30  miles  an  hour.  In  what  time  will  the 
second  automobile  overtake  the  first  ? 

46.  At  what  time  between  8  and  9  o'clock  do  the  hands  of 
a  clock  point  in  opposite  directions  ? 

47.  A  freight  train  goes  from  A  to  B  at  15  miles  per  hour. 
After  it  has  been  gone  4  hours  an  express  train  leaves  A  for  B, 
going  at  a  rate  of  45  miles  per  hour,  and  the  express  reaches 
B  ^  hour  ahead  of  the  freight.  How  many  miles  is  it  from 
AtoB? 

48.  In  traveling  a  certain  distance  a  train  going  45  miles 
an  hour  requires  5-J-  hours  less  time  than  an  automobile  going 
the  same  distance  at  27- miles  per  hour.  What  is  the  distance 
between  the  two  points  ? 


CHAPTER   XV 
APPLICATIONS   OF   GENERAL   SYMBOLS.     REVIEW 

STATEMENTS.      PHYSICAL  FORMULAS.      DERIVED 
EXPRESSIONS 

THE  GENERAL  STATEMENT  OF  A  PROBLEM 

178.  From  the  following  illustrations  it  will  be  seen  that 
when  the  given  numbers  of  a  problem  are  literal  quantities, 
the  statement  and  the  solution  result  in  a  formula  or  general 
expression  for  that  particular  kind  of  problem. 

Illustrations  : 

1.  If  A  can  mow  a  field  in  m  days,  and  B  can  mow  the 
same  field  in  p  days,  in  how  many  days  can  both  together 
mow  the  field  ? 

Let  x  =  the  number  of  days  both  working  together  require. 

Then  -  =  the  portion  of  the  work  both  together  can  do  in  1  day. 

x 

Also  —  =  the  portion  of  the  work  that  A  alone  can  do  in  1  day. 

m 

—  =  the  portion  of  the  work  that  B  alone  can  do  in  1  day. 
P 

Hence,  — h  —  =  the  portion  of  the  work  both  together  can  do  in  1  day ; 
m     p 

and  — |-  -  =  -  is  the  required  general  equation  for  the  condition. 

m     p     x 

Solving,  mx  +px  =  mp. 

(m  +p)se  =  mp. 

X=_™P_.    Result. 

m  +p 

This  expression  is,  therefore,  a  formula  for  finding  the  time 
in  which  two  men  whose  individual  ability  is  known,  can,  work- 

SOM.    EL.    ALG. — 11  161 


162      PRACTICAL  APPLICATIONS  OE  GENERAL   SYMBOLS 

ing  together,  accomplish  a  given  task.  By  substituting  in  this 
formula,  any  problem  involving  the  same  condition  can  be 
solved.     For  example : 

A  requires  4  days  to  do  a  certain  task,  and  B  requires  5  days 
for  the  same  work.  In  how  many  days  can  both  working  to- 
gether complete  the  work? 

Here  we  have  A's  time  alone  (or  m)  =  4 ;    B's  time  alone 

(or  p)  =  5.     In  the  formula  x  =  -^£-  =  ijA  =  ^  =  2f  days,  , 

v     i}  m+p     4  +  5      9         9 

the  time  in  which  both  together  can  do  it. 

2.  Divide  the  number  a  into  two  parts  such  that  m  times 
the  smaller  part  shall  be  contained  q  times  in  the  larger  part. 

Let  x  =  the  smaller  part. 

Then  a  —  x  =  the  larger  part. 

From  the  conditions, 

«^  =  g, 

mx 

and  a  —  x  =  mqx. 

mqx  +  x  =  a. 

(mq  +  l)x  as  a. 

x  s=  — - — ,  the  smaller  part  required. 
mq+  1 

Also,         a-x  =  a 2_  =  amg  +  a-a  =    «m<?    ?  the larger part 

m«+l  «g  +  l  ««  +  l       required. 

To  use  this  formula :  Suppose  we  are  required  to  divide  60 
into  two  parts  such  that  twice  the  smaller  part  shall  be  con- 
tained 7  times  in  the  larger  part.     We  have 

a  =  60,  x  =  — — —  =  4,  the  smaller  part  required. 

w»  =  2.  2-7  +  1        ' 

q  =  7.  f>A    o    <t 

a_x_  w  -  *  >  t  _  56   the  iarger  part  required. 

2-7  +  1 
And,  verifying,  56  +■  2(4)  =  7. 


GENERAL   STATEMENT  OF  A  PROBLEM  163 


Exercise  55 

1.  Divide  the  number  c  into  two  parts  such  that  m  times 
the  larger  part  shall  equal  n  times  the  smaller  part. 

2.  Divide  a  into  two  parts  such  that  the  sum  of   -th  of  the 

n 

larger  part  and  -th  of  the  smaller  part  shall  be  q. 
r 

3.  The  sum  of  two  numbers  is  s,  and  if  the  greater  number, 
g,  is  divided  by  the  less  number,  the  quotient  is  q  and  the 
remainder  r.     Find  the  numbers. 

4.  If  A  and  B  can  together  mow  a  field  in  t  days  and  B 
alone  can  mow  the  same  field  in  b  days,  find  the  number  of 
days  that  A  working  alone  will  require  to  do  the  work. 

5.  Show  that  the  difference  of  the  squares  of  any  two  con- 
secutive numbers  is  1  more  than  double  the  smaller  number. 

6.  m  times  a  certain  number  is  as  much  above  k  as  d  is 
above  c  times  the  same  number.     Find  the  number. 

7.  A  and  B  are  m  miles  apart,  and  start  to  travel  toward 
each  other.  If  they  start  at  the  same  time  and  A  goes  at  a 
rate  of  &  miles  an  hour  while  B  goes  at  the  rate  of  s  miles,  how 
far  will  each  have  gone  when  they  meet  ? 

8.  When  a  certain  number  is  divided  by  a,  the  quotient  is 
c  and  the  remainder  m.     Find  the  number. 

9.  The  front  wheel  of  a  wagon  is  m  feet  in  circumference, 
and  the  rear  wheel  n  feet  in  circumference.  How  far  has  the 
wagon  gone  when  the  rear  wheel  has  made  r  revolutions  less 
than  the  front  wheel  ? 

10.  A  and  B  can  together  build  a  barn  in  r  days,  B  and  C 
the  same  barn  in  s  days,  and  A  and  C  the  same  in  t  days.  In 
how  many  days  can  each  alone  build  it  ? 


164     PRACTICAL   APPLICATIONS   OF   GENERAL   SYMBOLS 
THE  USE  OF  THE  LABORATORY  FORMS  OF   PHYSICS 

179.  Density  is  defined  as  mass  per  unit  of  volume,  and  may 

M 

be  calculated  by  the  formula  D—^,  in  which  D  is  the  density 

of  the  body  under  examination,  M its  mass,  and  Fits  volume. 
Volume  is  another  name  for  cubical  contents.  The  mass  of  a 
body  may  be  found  by  weighing  it,  and  the  formula  for  density 

W 

may  be  written  D  =  — ,  in  which  W  is  the  weight  of  the  body 

to  be  considered. 

Exercise  56 

1.  A  block  of  iron  3  feet  6  inches  long,  2  feet  wide,  and  6 
inches  thick  weighs  1706.25  pounds.  Calculate  its  density  in 
pounds  per  cubic  foot. 

2.  The  density  of  water  is  approximately  62.5  pounds  per 
cubic  foot.     What  is  the  volume  of  a  ton  of  water  ? 

3.  A  cubical  block  of  wood  9  centimeters  on  an  edge  has  a 
mass  of  371.79  grams.  Calculate  its  density  in  grams  per 
cubic  centimeter. 

4.  A  block  of  iron  having  an  irregular  cavity  weighs  3.265 
kilograms.  When  the  cavity  is  filled  with  mercury,  the  whole 
weighs  3997  grams.  The  density  of  mercury  being  13.6  grams 
per  cubic  centimeter,  calculate  the  volume  of  the  cavity. 

5.  The  density  of  a  certain  substance  being  a  grams  per 
cubic  centimeter,  calculate  the  mass  of  this  substance  necessary 
to  fill  a  vessel  of  1000  cubic  centimeters'  capacity. 

180.  To  change  a  reading  on  a  Centigrade  thermometer  to  a 

corresponding  reading  on  a  Fahrenheit  thermometer  use  is 

made  of  the  formula 

_F=.f  0+ 32, 

In  which  F  is  the  reading  in  degrees  on  the  Fahrenheit  scale, 
and  C  the  reading  in  degrees  on  the  Centigrade  scale.     To 


LABORATORY  FORMULAS  OF  PHYSICS  165 

change  Fahrenheit  readings  to  Centigrade  readings  we  use  the 
formula 

C  =  f(F-  32). 

Exercise  57 

1.  Change  the  following  Centigrade  readings  to  correspond- 
ing Fahrenheit  readings  :  20°;  40°;  0°;   -15°;   -273°;   -180°. 

2.  Change  the  following  Fahrenheit  readings  to  correspond- 
ing Centigrade  readings  :  212°;  32°;  70°;  -10°. 

3.  At  what  temperature  would  the  reading  on  a  Centigrade 
thermometer  be  the  same  as  the  reading  on  a  Fahrenheit 
thermometer? 

4.  The  Centigrade  scale  is  marked  0°  at  the  freezing  point 
of  water,  and  100°  at  the  boiling  point  of  water.  The  Reaumur 
thermometer  is  marked  0°  at  the  freezing  point  of  water  and  80° 
at  the  boiling  point.  Prepare  (1)  a  formula  for  changing  Reau- 
mur readings  to  Centigrade  readings,  and  (2)  a  formula  for 
changing  Centigrade  readings  to  Reaumur  readings. 

181.   In  the  figure,  we  have  a  straight  bar  whose  weight  is 

__  ,m m  to  be  neglected.     The  bar  is  sup- 

——d- \~jf~  !      ported    at  the  point   C  arid   is 

1      acted  upon  by  the  several  forces, 


,«// 


f     C 


t»  /?  /'?  /">  etcv  *n  tne  directions 
indicated  by  the  arrows.  These 
forces  tend  to  cause  rotation  of 
the  bar  about  the  axis  C.  The  tendency  of  a  force  to  produce 
rotation  is  called  its  moment,  and  the  moment  is  calculated  by 
multiplying  the  magnitude  of  the  force  by  the  distance  of  its 
point  of  application  from  the  axis.  Moments  tending  to  rotate 
a  body  clockwise  are  given  a  positive  sign ;  those  tending  to 
produce  rotation  in  the  opposite  direction  are  given  a  negative 
sign.  In  order  that  a  body  under  the  influence  of  moments 
may  be  in  equilibrium,  i.e.  stationary,  the  algebraic  sum  of  the 


166   PRACTICAL  APPLICATIONS  OF  GENERAL  SYMBOLS 

moments  acting  on  it  must  be  0.     Thus  in  the  figure,  if  the 
bar  is  in  equilibrium, 

f'd"  +  f'"d'"  -fd  -f'd'  -f""d""  =  0, 
or,  better : 

(f'd"  +  /'"<*'")  -  (fd  +fd'  +f""d"")  =  0. 
(/'»/"»/"'»  etc.,  are  read  "/prime,"  "/second,"  "/third"  etc.) 

Exercise  58 

1.  Two  weights  of  4  and  12  pounds  respectively  are  bal- 
anced on  a  bar,  the  distance  from  the  support  to  the  further 
weight  being  6  feet.  What  is  the  distance  from  the  support 
to  the  nearer  weight  ? 

2.  A  bar  11  feet  long  is  in  equilibrium  when  weights  of 
15  and  18  pounds  are  hung  at  its  ends.  Find  the  distance  of 
the  point  of  support  from  each  end. 

3.  If    weights    are    distrib-  , 


— v: — ,o-- 


ED  EI 


uted  upon  a  bar  as  in  the 
figure,  where  must  a  weight 
of  40  pounds  be  placed  to  keep 
the  bar  in  equilibrium  ? 

4.  A  10-pound  weight  hangs  at  one  end  of  *a  12-foot  bar, 
and  a  15-pound  weight  hangs  at  the  same  side  of  the  support- 
ing point,  but  2  feet  nearer  it.  If  a  40-pound  weight  at  the 
other  end  keeps  the  bar  in  equilibrium,  at  what  distances  from 
the  ends  is  the  point  of  support  located  ? 

THE  VALUE  OF  ANY  ONE  ELEMENT  OF  A  FORMULA  IN  TERMS  OF  THE 
OTHER  ELEMENTS 

182.  The  statement  of  a  mathematical  law  by  means  of  a 
formula  always  gives  an  expression  for  the  value  of  the  par- 
ticular element  to  which  the  law  refers.  By  transpositions 
and  divisions  we  are  able  to  derive  from  any  formula  another 
expression  or  formula  for  any  oue  of  the  other  elements. 


TRANSFORMATION   OF   FORMULAS  167 

Illustrations : 

1.  Given  the  formula,  R  =    &s    .     Derive  a  formula  for  s. 

9  +  s 

Clearing  of  fractions,  (g  +  s)  B  =  gs. 

Multiplying,  Bg+  Bs  =  gs. 
Transposing,  Bs  —  gs  =  —  Bg. 

Dividing  by  —  1,  gs  —  Bs  =  Bg. 

Collecting  coefficients,  (g  —  B)s  =  Bg. 

Dividing  by  (g  —  B) ,  s  =  — 2-,  the  required  formula  for  s. 

g  —  B 

2.  Given  the  formula,  I  =  a  +  (n  —  1)  d.     Find  an  expression 
for  n.  * 

l  =  a+  (n-l)d. 
Multiplying,  I  =  a  +  nd  —  d. 

Transposing,  I  —  a  +  d  =  nd. 

Dividing  by  d,  n=    ~  a  +    ,  the  required  formula  for  n. 

d 


THE   TRANSFORMATION  OF  FORMULAS 

Exercise  59 
PHYSICAL  FORMULAS 

1.  Given  v  =  at,  find  the  value  of  t  in  terms  of  a  and  v. 

2.  Given  S  =  %  g  (2 1  —  1),  find  a  formula  for  t  in  terms  of  s 

and  g. 

3.  Given  C  = ,  derive  a  formula  for  S  in  terms  of  E, 

f*9 


b,  P,  C,  and  R. 

4.   Giv 
p,  and  p' 


4.  Given  -=--{-—,  find  an  expression  for  each  element,  /, 


168      PRACTICAL  APPLICATIONS  OF  GENERAL  SYMBOLS 

5.  Given  F  =  f  C  +  32,  derive  a  formula  for  C  in  terms  of  F. 

6.  Given  F  =  — ,  v  =  gt,  and  w  =  .Fs,  find  a  value  for  to 
in  terms  of  m,  g,  and  s. 

7.  Given  ^1  =  ^M,  derive  expressions  for  2\  and  v2- 


MISCELLANEOUS  FORMULAS 

8.  Given  0=2  irR,  obtain  a  formula  for  M. 

9.  Given  /  =  a  -f  (n  —  1)  d,  derive  an  expression  for  a  in 
terms  of  I,  n,  and  d. 

10.  Given  8  =  ^  (a  ■+■  Q>  find  the  value  of  a  in  terms  of  Z,  n, 
and  & 

11.  Given  T=nR(R  +  L),  find  £  in  terms  of  the  elements 
involved. 

12.  Given  S  =  r  ~a,  find  each  element  in  terms  of  the 
r- 

others. 


13.  Given   d=-±- -1,  find  a  formula  for  S. 

n(n  —  l) 

14.  Given  a=p-\-prt,  find  each  element  in  terms  of  the 
others. 


GENERAL  REVIEW 

Exercise  60 


1.  Simplify  2a-  [3-2{a-4(a-a  +  l)}]. 

2.  Solve  (a>-l)(«  +  3)-2(«-l)(3»+l)  =  (3-«)(2+5a). 

3.  Showthatfl-f-4^— ^-Y-^- i_+lVl=0. 

\       a  +  1     a-f-3/\a  — 3     a  — 1       / 


GENERAL  REVIEW 


169 


8 

or     a 

rl      1 

a     2 

M 

la     2) 

a      azJ 

4.   Simplify 


5.  Factor  a5-S(a2-x2)-a3xi. 

6.  What  number  added  to  the  numerators  of  the  fractions 

-  and  -,  respectively,  will  make  the  results  equal?     Is  there 
b  d 

an  impossible  case  ? 

7.  Factor  6  x3  +  6  a2x  -  37  ax2. 


9.   Solve 


+ 


l-2a;     l-4a;     l-3z 

10.  Factor  (a2  -  9)  (a  +  2)  -  a  -  (4  +  a)  (3+  a)  -  3. 

11.  Solve  ^±^ S_=^-m 

3         a  +  ra        3 

12.  By  three  different  methods  factor  (sc2  —  6)2  —  a£ 

5 


13.   Solve  JtlJfl-lL 
2,-1     4^,-1     37 


6(1-2/) 


14.  For  what  value  of  aiscc4  —  3a^  +  2a^  +  21a;  +  3a  divis- 
ible by  x2  +  x-  3  ? 

15.  The  sum  of  the  numerator  and  the  denominator  of  a 
certain  fraction  is  39.  If  3  is  subtracted  from  both  numerator 
and  denominator,  the  result  is  \.     Find  the  original  fraction 

16.  Factor  225  -  4  x2  (9  +  x)  (9  +  x). 

17.  Find  the  H.C.F.  and  the  L.  CM.  of  x4  -  ax*  -  2  a2x>, 
2ar5-2a2a;,  and  3  ar>  + 12  aa2  +  3  a2x. 


170    PRACTICAL  APPLICATIONS  OF  GENERAL  SYMBOLS 

18.  Simplify    (C-±±-C-=i)  +  (c±A  +  c^±).      ■ 

19.  Solve  and  verify  2-^=±  +  ^  =  *JL=1 . 

3        7  2s  +  2         28 

20.  Prove  that  -i- 1 —  =  cd- 

1+i  1-1     od-I 

cd  cd  cd 

21.  Simplify    (a-l)2-4^r_3 21     (a-3} 

p     J     (a+2)2-l      [a  +  3     a-3j      a-U 

22.  Solve 


2 

15 


(M-.l     a  +  2     a  +  3 

23.  A  boy  has  a  dollar,  and  his  sister  has  28  cents.  He 
spends  three  times  as  much  as  she  spends,  but  has  left  four 
times  as  much  as  she  has  left.     How  much  did  each  spend  ? 

24.  What  is  the  value  of  m  if  3m  +  5n  =  1  when  n  =  - 1. 

m  —  n 

■2  S_x  _7 

25.  Solve  £_  +  -* -— 4— J- 

f(a^l)      1(1+,)      1-1 

26.  Prove  that  the  sum  of  any  five  consecutive  numbers 
equals  5  times  the  middle  one. 

27.  Factor  (a2 -{-5  a- 10)2  +  2(a2  +  5  a-  10)a - 8  a2. 

28.  Find  three  consecutive  numbers  such  that  if  the  second 
and  third  are  taken  in  order  as  the  digits  of  a  number,  this 
number  will  be  7  more  than  four  times  the  sum  of  the  1st  and 
3d  given  numbers. 

29.  If  a  =  2,  b  =  3,  and  c  =  —  4,  find  the  value  of 

(a2  +  c*)(3  a  -  c)V(7a  +  c)(4  6-a). 


30.    Simplify   - 
x 


GENERAL   REVIEW 
1 


171 


1   Sx2-* 

i 5 1  + 

3     3a  +  l  I 


J — 2x 


2x 


2x 


+3 


31.  At  what  time  between  8  and  9  o'clock  are  the  hands 
of  a  watch  5  minutes  apart  ? 

32.  Find  the  H.  C.  F.  and  the  L.  C.  M.  of 

aj»-4,  8-a?8,  -4+4a-a2,  (2-oj)*,  and  (a?-2)(l-a?). 

/>»2 ^yj  ~, 

33.  Solve  and  verify   -^ (-  - 


4^  —  m      4     2a?  +  m 


34. 


Simplify  g  +  i+^)+j 


a?+l  .  x(x-l)+l~\  .     a2 


cc2  »3— 1  a^— 1 


35.  What  value  of  x  will  make  (4  x  +  3)  (3  x  —  1)  equal  to 
(6o>  +  5)(2»-l)? 

36.  Simplify     (6-c)(c-a)(a-5)      5-c     c-a      g-6, 

J      (&4-c)(c  +  a)(a  +  &)      &  +  c     c  +  a     a +6 

37.  By  what  must   x4-}-2xs-\-Sx2-\-4:X-{-5   be  divided  to 
give  a  quotient  of  ar*  +  4  a?  -}- 14  and  a  remainder  of  44  a?  +  47  ? 

8^-1 


38.    Simplify   _. 


1 


1- 


l+» 


39.  Show  that  if  a  =  -  1, 

i(l-"H  1('-)K1+-) 
SFFi "  JF+3R 

40.  Factor  mV  -  mV  -  27  »V  +  27. 

41.  Solve   fg->  +  c)2-CT  =  a-3ca;. 

n  a 

42.  Factor  6  a2  — 6np  — (4n  — 9p)a. 


172      PRACTICAL  APPLICATIONS  OF  GENERAL  SYMBOLS 

43.  The  head  of  a  certain  fish  weighs  h  pounds,  the  tail 
weighs  as  much  as  the  head  and  -J  the  body,  and  the  body  as 
much  as  the  head  and  tail.  What  is  the  weight  of  the  fish 
in  terms  of  h  ? 

3a2-3al 


44.    Simplify 


I    6a?-6 


4a 


a  +  1 
48^~ 


45.  In  how  many  years  will  s  dollars  amount  to  a  dollars 
at  r  per  cent,  simple  interest  ?• 

46.  Solve  for  m :  (2m  +  2)(3  b  —  c)+2ac  =  2c(a  —  m). 

47.  If  a  certain  number  of  wagons  is  sold  at  $80  each, 
the  same  amount  is  received  as  when  10  less  are  sold  at  $  100 
each.     How  many  are  sold  in  each  case  ? 

48.  Find  the  value  of 

\a?  a  J  a?  cr  +  1  a2— 1  m 

49.  Factor  a2  +  a2 -2(1  -ax)—  (x  +  a). 

50.  Two  bills  were  paid  with  a  10-dollar  bank  note.  One 
bill  was  25  %  more  than  the  other,  and  the  change  received 
was  -i-f  the  smaller  bill.     Find  the  amount  of  each  bill. 

51.  Solve  for  s: 

(s  -  a)  (s  +  b)  -  (s  +  a)(s  -  b)  -  2(a  -  b)  =  0. 

52.  Factor  aJ6  -  Xs  —  2  (3  a?  +  4). 

53.  Solve  for  n,     I  =  (n  —  1)  (a  —  I)  +  a. 

54.  An  orderly  is  dispatched  with  an  order,  and  3  hours 
after  he  leaves  a  second  orderly  is  sent  after  him  with  instruc- 
tions to  overtake  the  first  in  6  hours.  To  do  this  he  must 
travel  4  miles  an  hour  faster  than  the  first  traveled.  How 
many  miles  an  hour  does  each  travel  ? 

55.  Factor  50  x  +  2  x5  -  38  Xs. 


CHAPTER  XVI 

SIMULTANEOUS  LINEAR  EQUATIONS.    PROBLEMS 

183.   If  x  and  y  are  two  unknown  quantities  and  their  sum 
equals  7,  we  may  write 

x  +  y  =  7. 

Clearly,  an  unlimited  number  of  values  of  x  and  y  will  satisfy 
this  equation.     For  example : 


If  x  =  2, 

y  =  5. 

If*  =  l, 

y  =  6. 

If  x  =  0, 

y  =  7.    • 

11*=  - 

1, 

2/  =  8. 

Iix=  - 

2, 

y  =  9,  etc. 

184.  Such  an  equation  in  two  unknown  quantities,  satisfied 
by  an  unlimited  number  of  values  for  the  unknown  quantities, 
is  an  indeterminate  equation. 

185.  If,  however,  we  have  with  this  equation  a  second  equa- 
tion stating  a  different  relation  between  x  and  y}  as 

x  -  y  =  3, 
then  the  pair  of  equations, 

x  +  y  =  7, 

x  -  y  =  3, 
is  such  that  each  is  satisfied  only  when  x  =  5  and  y  =  2. 
For  x  +  y  =  5  +  2  =  7, 

And  x-y  =  5-2  =  3. 

173 


174  SIMULTANEOUS   LINEAR  EQUATIONS 

No  other  values  of  x  and  y  will  satisfy  this  pair  of  equations. 
Hence, 

186.  Simultaneous  equations  are  equations  in  which  the  same 
unknown  quantity  has  the  same  value. 

187.  A  group  of  two  or  more  simultaneous  equations  is  a 
system  of  equations. 

188.  Two  possible  cases  of  equations  that  the  beginner  may 
confuse  with  simultaneous  equations  must  be  carefully  noted. 

(a)  Inconsistent  Equations. 

Given :  x  +  y  =  9, 
x  +  y  =  2. 

It  is  manifestly  impossible  to  find  a  set  of  values  for  x  and  y  that 
shall  satisfy  both  given  equations.     The  equations  are  inconsistent. 

(b)  Equivalent  Equations. 

Given  :        x  +  y  —   4, 
3a +  3^  =  12. 

If  the  first  equation  is  multiplied  by  3,  it  becomes  the  same  as  the 
second  equation,  and  every  set  of  values  that  satisfies  the  second  satis- 
fies the  first  as  well.     The  equations  are  equivalent. 

189.  For  a  definite  solution  of  a  pair  of  simultaneous  equa- 
tions we  must  have  a  different  relation  between  the  unknown 
quantities  expressed  by  the  given  equations. 

Equations  that  express  different  relations  are  independent 
equations. 
• 

190.  Simultaneous  equations  are  solved  by  obtaining  from 
the  given  equations  a  single  equation  with  but  one  unknown 
quantity.  This  process  is  elimination.  Each  of  the  three 
methods  of  elimination  in  common  use  should  be  thoroughly 
mastered. 


ELIMINATION  BY  SUBSTITUTION  175 
ELIMINATION  BY  SUBSTITUTION 

Illustration : 

(5x  +  2y  =  ll,  (1) 

Solve  the  equations :    {  ^         .  ' 

H                  {3x  +  ±y  =  l.  (2) 


l-4y 
3 


From  (2), 

Substituting  in  ( 1 ) ,       5 1 1-4y>\  +  2 y  =  11. 

From  which  5-20j  +  2  y  =  n. 

o 

Clearing,  5  -  20  y  +  6  y  =  33. 

-  14  y  =  28. 

y=-2. 
Substituting  in  (1),  5x  +  2(-  2)  =  11. 

5x-4  =  ll. 
x=    3. 

l  Result. 
y  =  -  2.  J 

Check : 

Substituting  in  (1),        5  (3)  +  2  (-  2)  =  15  -  4  =  11. 

Substituting  in  (2),        3  (3)  +  4  (-  2)  =  9  -  8  =  1. 

From  the  illustration  we  have  the  general  process  for  elimi- 
nation by  substitution : 

191.  From  one  of  the  given  equations  obtain  a  value  for  one 
of  the  unknown  quantities  in  terms  of  the  other  unknown  quan- 
tity.    Substitute  this  value  in  the  other  equation  and  solve. 

The  method  of  substitution  is  of  decided  advantage  in  the 
solution  of  those  systems  in  which  the  coefficients  of  one  equa- 
tion are  small  numbers.  In  later  algebra  a  knowledge  of  this 
method  is  indispensable. 

In  applying  this  process  of  elimination  care  should  be  taken 
that  the  expression  for  substitution  is  obtained  from  the  equa- 
tion whose  coefficients  are  smallest.  The  resulting  derived 
equation  will  usually  be  free  from  large  numbers. 


176  SIMULTANEOUS  LINEAR   EQUATIONS 

ELIMINATION  BY  COMPARISON 

Illustration : 

5x  +  2y  =  9,  (1) 


Solve  the  equations  :  , 

'2a5  +  32/  =  8.  (2) 

From(l),     x=9~2y'  From  (2),       x-*^^- 


2 


By  Ax.  5,  9-21  =  8-3_y< 

6  2 

Clearing,  2  (9  -  2  y)  =  5  (8  -  3  y). 

18  -  4  y  =  40  -  15  y. 

11  y  =  22. 

2/ =  2. 

Substituting  in  (1),     5  x  +  2  (2)  =  9. 


Hence,  *,  _ 


X  =  1,1 

2/ =  2.  J 


Check : 

Substituting  in  (1),        5  (1)  +  2  (2)  =  5  +  4  =  9. 
Substituting  in  (2),        2  (1)  +  3  (2)  =  2  +  9  =  8. 

In  general,  to  eliminate  by  comparison : 

192.  From  each  equation  obtain  the  value  of  the  same  unknown 
quantity  in  terms  of  the  other  unknown  quantity.  Place  these 
values  equal  to  each  other  and  solve. 

The  method  of  comparison  is  particularly  adapted  to  those 
systems  of  simultaneous  equations  in  which  the  coefficients  are 
literal  quantities. 

ELIMINATION  BY  ADDITION  OR  SUBTRACTION 

Illustrations : 

1.    Solve  the  equations  :   \Sx-±y=5> 

H  l5a>  +  3y  =  18.  (2) 

Choosing  the  terms  containing  y  for  elimination,  we  seek  to  make  the 
coefficients  of  y  in  both  equations  equal  ;  for,  if  these  coefficients  were 


ELIMINATION  BY  ADDITION  OR  SUBTRACTION      177 

equal,  adding  the  equations  would  cause  y  to  disappear.  The  L.  C.  M. 
of  the  coefficients  of  y,  3  and  4,  is  12.  Dividing  each  coefficient  into  12, 
we  obtain  the  multipliers  for  the  respective  equations  that  will  make  the 
coefficients  of  y  the  same  in  both. 

Multiplying  (1)  by  3,  9  x  -  12  y  =  15  (3) 

Multiplying  (2)  by  4,  20  x  +  12  y  =  72  (4) 

Adding  (3)  and  (4),  29  x  =  87 

x  =3. 

Substituting  in  (1),  3  (3)  -  4  y  =  5.     T 

9-4,  =  5.     Hence,*  =  3,l     Regult 

,  =  1.  y  =  1*i 

It  is  to  be  noted  that  the  signs  of  y,  the  eliminated  letter,  being  unlike, 
the  process  of  addition  causes  the  y-term  to  disappear. 


2.   Solve  the  equations 

I2x-3y  =  l,                                (1) 
\3x  +  7y  =  13.                              (2) 

Multiplying  (1)  by  3, 
Multiplying  (2)  by  2, 
Subtracting  (4)  from  (3), 

6  x    -  9  y  =  3.                                     (3) 
6  x  +  14  y  =  26.                                     (4) 
-  23  y  =  -  23. 
2/  =  l. 
2s-3(l)=l.     tt 

2a- 3  =  1.     Hence'*  =  M  Result. 

x=2,         y=1-i 

Substituting  in  (1), 

In  this  example  since  the  signs  of  the  sc-term  are  like,  the  equations 
are  subtracted  to  eliminate  the  sc-term  having  the  same  coefficients. 

In  general  to  eliminate  by  addition  or  subtraction : 

193.  Multiply  one  or  both  given  equations  by  the  smallest  num- 
bers that  will  make  the  coefficients  of  one  unknown  quantity  equal. 
If  the  signs  of  the  coefficients  of  the  term  to  be  eliminated  are 
unlike,  add  the  equations  ;  if  like,  subtract  them.  The  necessary 
multipliers  for  the  elimination  of  a  letter  may  be  found  by  dividing 
each  coefficient  of  that  letter  into  the  lowest  common  multiple  of  the 
given  coefficients  of  that  letter. 

The  method  of  addition  or  subtraction  is  of  advantage  when 
it  is  desirable  to  avoid  fractions,  and  is  used  in  the  solutions 
of  systems  involving  more  than  two  unknown  quantities. 

SOM.    EL.    ALG. 12 


178  SIMULTANEOUS  LINEAR  EQUATIONS 


Exercise  61 


Solve: 


1.  x  +  y  =  3,  10.   3s  +  7y  =  -8, 

3x  +  2y  =  7.  s  +  y  =  0. 

2.  a;-2/  =  4,  11.    5x  +  3y  =  10, 
2x  +  3y  =  13.  Sx-5y  =  16. 

3.  4  x  +  3  y  =  7,  12.    7#  —  3  =  4y, 
3<c-2/  =  2.  o;  +  2/-2  =  0. 

4.  6cc-5?/  =  7,  13.    5#  +  8?/-2  =  0, 
a;  +  92/  =  ll.  7y  +  72-3x  =  0. 

5.  3o;  +  4z  =  2,  14.    10m  +  llw  =  32, 
4  a;- 2  =  9.  15  m +  31  =  23™. 

6.  3s-5*  =  13,  16.    3y-±x  =  77, 
2s  +  7t  =  -12.  6y  +  x  =  l. 

7.  3x  +  y  =  -10,  16.    5s  +  3*  =  112, 
2a?-5y  =  -l.  4*- 5  s -49  =  0. 

8.  4v-3w  =  0,  17.    5^4-13^  =  4, 
2  v  +  5  w  =  26.  7  n  + 19  v  =  4. 

9.  5«  +  7?/  =  24.  18.    12p-7Z-9  =  0, 
x-y=0.  15  t -7 p  =-300. 

SIMULTANEOUS  LINEAR  EQUATIONS  CONTAINING  THREE  OR  MORE 
UNKNOWN  QUANTITIES 

194.  A  system  of  three  independent  equations  involving 
three  unknown  numbers  is  solved  by  a  repeated  application 
of  the  process  of  elimination.  The  method  of  addition  or  sub- 
traction is  most  commonly  used  with  systems  having  three  or 
more  unknown  quantities. 


THREE  OR  MORE  UNKNOWN  QUANTITIES 


179 


Illustrations : 


-3a;-22/  +  3z  =  ll, 

(1) 

1.   Solve  the  equations : 

2x  —  3y  +  2z  =  9, 

(2) 

.3#  +  52/  +  4z  =  6. 

(3) 

Multiply  (1)  by  8, 

9x-6y+9z  =  S3 

(4) 

Multiply  (2)  by  2, 

4x-6y  +  4z  =  18 

(•5) 

Subtract  (5)  from  (4), 

5x            +5^  =  15 

(6) 

Divide  by  5, 

x  +  z  =  3. 

(7) 

Multiply  (2)  by  5, 

10  x  -  15  y  +  10  z  =  45 

(8) 

Multiply  (3)  by  3, 

9x  +  15^  +  12  0  =  18 

(9) 

Adding  (8)  and  (9), 

19  x              +  22  z  =  63 

(10) 

Multiply  (7)  by  19, 

19  a;              +  19  z  =  57 

Subtracting, 

30  =  6 
0  =  2. 

Substituting  in  (7), 

x  +  (2)  =  3,  x  =  1.        x-  1,     ] 

Substituting  in  (1), 

3(1) -2^  +  3(2)  =  11.      y=-l,l 
y=-l.    2=2.     J 

Result. 

r*+2y=l, 

(i) 

2.   Solve  the  equations :   \3  x  —  z  =  —  8, 

(2) 

[y  +  2z=0. 

(3) 

Multiply  (1)  by  3, 

3x  + 6y =3 

(*) 

Subtracting  (2), 

Multiply  (5)  by  2, 
Subtract  (3)  from  (5), 


6y  +     0  = 
12  y  +  2  0  = 

=  11 
=  22 

y +20  = 

=    0 

Substituting  in  (3), 
Substituting  in  (1), 


11  y  =22 

y  =    2. 

(2) +20  =  0,     0=-l. 
x  +  2(2)=l,     x=-3. 


53 


(5) 


Result. 


195.  It  is  important  to  note  that  if  three  unknown  quanti- 
ties are  involved,  three  independent  equations  must  be  given. 
Similarly,  with  four  unknown  quantities,  four  independent 
equations  are  necessary,  etc. 


180 


SIMULTANEOUS  LINEAR   EQUATIONS 


From  the  illustrations  we  may  state  the  general  process : 

196.  Eliminate  one  unknown  quantity  from  any  convenient  pair 
of  equations,  and  the  same  unknown  quantity  from  a  different  pah- 
of  equations.  Solve  the  resulting  equations  by  any  of  the  methods 
already  given. 


Exercise  62 


Solve : 

1.  x  +  y  +  z  =  9, 
x-y  +  z  =  3, 
x  —  y  —  z  =  l. 

2.  x  —  y  -fz  =  5, 
y-x  +  z  =  -l, 
z  +  y  +  x  =  19. 

3.  x  +  y-z  =  4, 

y  —  x  —  z  —  — 10, 

z  +  y  —  x  =  0. 

4.  u  +  2v  +  w  =  4:, 
u  —  v  +  2w  =  2, 
2u-\-v  —  w  —  2. 

5.  r-f 2s  +  3£  =  14, 

2  r  -  s  + 1  =  3, 
3r-2s-*  =  -4. 

6.  3x  —  2y—  z=—  8, 
5x  —  3y  +  z  =  l, 
2x  +  7y  +  3z=38. 

7..    4  a;  —  3  y  4-  5  z  =  —  15, 
4z-3z  +  2?/  =  3, 
4  2/ 4- 7  a;  —  3z  =  4. 


8.  10m  +  3n-2jt>  =  22, 
3m  — 5n4-7p  =  — 1, 
8m-9n-5jp  =  21. 

9.  a-hz=10, 
2/-z  =  2, 
«-2/  =  2. 

10.  2  #  +  2/  =3, 
3z-x  =  -3, 
±y-3z  =  -12. 

11.  a?4-2/  +  z  =  0, 
5x-3y  =  50, 
2z  +  y  =  -20. 

12.  y  =  5  x  +  17, 
3^-22/4-37  =  0. 
71  =  -  7  *  4-  2  x. 

13.  a?4-2/  +  z=24, 
x  +  y  +  u  =  25, 
x  +  z  +  u  =  26, 
y  +  z+u  =  27. 

14.  a?4-y=18, 
2/  4-  *  =  14, 
z  +  w  =  10, 
w  4-  w  =  6, 
a  4-  u  =  12. 


FRACTIONAL  EQUATIONS  181 


FRACTIONAL  FORMS  OF  SIMULTANEOUS  LINEAR  EQUATIONS 

(a)  When  the  Unknown   Quantities  occur  in  the  Numera- 
tors of  the  Fractions 

197.  Systems  of  simultaneous  linear  equations  in  which  the 
fractions  have  unknown  quantities  in  the  numerators  only, 
are  solved  by  first  clearing  of  fractions  and  then  eliminating 
by  either  of  the  three  methods. 

Illustration:  [V  -  *     1_^~  4  m 

~2~  3     '  W 


Solve  the  equations : 


3  4  w 


From  (1),  By  -Bx-6  =  2x-S. 

Simplifying,  5  x  -  3  y  =  2.                                     (3) 

From  (2),  4  (2y  +  1)  -  3  (a  +  3)  =  0. 

Simplifying,  3  x  —  8  y  =  —  5.                                (4) 

From  (3),  x  =  ?JL±2.  From  (4),        x  =  ^^ . 

o  3 

Therefore,  Sy±2  =  Sy-5t  (5) 

5  3 

Solving  (5),  y  =  l,l     Regult> 

Substituting  in  (1),  x  =  1>  J 


198.  Extraneous  Roots.  In  fractional  simultaneous  equa- 
tions we  may  reject  any  solution  that  does  not  satisfy  both 
given  equations. 

Exercise  63 

Solve: 


1. 

*  ,  y  - 

3^4 

6, 

4^~2 

7. 

2. 

X 

5~ 

y  _ 

3 

•■  — 

X 

4 

.y  - 

7 

:2. 

3, 


3x 
2 

T  3  " 

=  17, 

2x 
3 

■3 

=  13. 

4<c 
3 

5 

=  12, 

Sx 
4 

T  2  ' 

=  34. 

182  SIMULTANEOUS  LINEAR  EQUATIONS 

x+3     y+l+  q     x+1     2-y_     5 

x-2     y-2_i  3a?-l     y-S_     1 


8. 


2  3  4 


6>    2^fl  +  3^1  =  6>  10     ?_4  =  |+6, 

3a;  — 1      2y  — l_g  a;  +  y      a;  —  2  y  =  35     a; 

2  +      5  '  10    *"      8  2      6* 

5y-2      a;-9=5  3s  +  l      4y  +  l  =     1 

3  2         2'  3  2  2' 
3a?~l     y  +  4     1=0  2a?-l     4y-l^l 

2  3         6'  2  32* 


2  3     ~4'  iL_?.4-5 5-5 

a2J-l_y+l==_5.  '       3     +     4     ~8? 

3  2             6*  5  (x  +2/)  =  a;  —  y. 

2y-a;-3     y-2x-3_, 

13.             j  -          -4, 

4y-3a;-3     2y-4a;  +  9_0 

14.     *  +  2>+*  =  10,  15.    ^-^  +  ^  =  0, 

2T3     4  3        2        6' 

»     y_z_8  4a;  .  2y     3z_l 

3     2     8~   '  3        3        2       2' 

a:_y     2_o  5a;     5y     4z_7 
6     5  +  4~   ' 

16.    |  +  |  =  2,  17. 

2  3 

2>+*=4. 

3  2 


4 

6 

•     y__  * 
2     3     12' 

18 

a;      z_  1 
3     4~24' 

2/      *_  *  . 
2     3     12 

a; 
3" 

y_    11 

2         72 

2/ 
4 

2       1 
"3     48' 

a; 

2 

z  _  5 
"5     24* 

FRACTIONAL  EQUATIONS 


183 


(b)  When  the   Unknown   Quantities    occur   in   the   Denom- 
inators  of   the    Fractions,   giving    Simultaneous  Linear 

Equations  in  -  and  — 
x  y 

199.   A  solution  of  this  type  of  simultaneous  equations  is 

obtained  by  considering  the  unknown  quantities  to  be  -  and  -, 

x  y 

and  the  process  of  elimination  is  carried  through  without  clear- 
ing of  fractions.     Much  difficulty  is  avoided  by  this  method. 

Illustrations : 

f3     2==_31 
x     y         40 ' 
5_10  =  11# 
x      y       8 


1.   Solve  the  equations : 


I 


Multiplying  (1)  by  5, 
Adding  (2), 

Dividing  by  20, 
Clearing  of  fractions, 


Substituting  in  (2) ,   -  -  -  —  =  — 


16  , 

10 

31 

V 

— 

—  — 

-  — 

X 

y 

8 

5_ 

10 

11 

X 

y 

8 

20 

20 

X 

8 

1 

1 

X 

Y 

X 

=- 

-8. 

11 

10 

16 

(1) 

(2) 
(3) 


8'  -so  =  16y<  y=-5- 


Hence,  x 

y 


::;:) 


Result. 


200.  If  the  coefficients  of  x  or  y  are  greater  than  1,  the 
equations  may  be  changed  in  form  by  multiplying  each  equa- 
tion by  the  L.  C.  M.  of  those  coefficients.  The  resulting  equa- 
tions will  be  of  the  same  form  as  the  equations  just  solved. 


2.  Solve  the  equations :  . 


2a;     3^ 
4a;     6y 


(i) 

(2) 


Multiplying  (1)  by  6 

i£  +  l?=_12. 

- 

Multiplying  (2)  by  12,      ?!  +  —  =  12. 
x       y 

From  the  system  (3) 

and  (4), 

jc  =  -  and  y  =  — 

1 

2         * 

3' 

Exercise  64 

Solve: 

,     1  ,  3     o 

3     5_7 

5        2  _11 

1.   -  +  -  =  2, 
x     y 

6. 

x     y     6' 

11. 

2a;     Sy      6' 

?-§  =  2. 

5_6  =  7 

9  2/  —  8  x  =  xy. 

x     y 

a;     2/     2 

12. 

1,1     17 

1                  ~  TrT  J 

A    2,3     ., 

1        1    _  5 

a:     y      z     12 

2.    -  +  -  =  1, 
x     y 

7. 

2a;     Sy     12' 

1     1,1      5 

1          —  -i  r>> 

4_3_1 

1        1  ___  1. 

a;     2/      z      ** 

a     2/     2 

3a;     2y         12* 

x     y      z         12 

„    3.5     o 

1        1.7 

3.    -  +  -  =  2, 
a  '  2/ 

8. 

3a;     2y         18' 

13. 

?-?+!=? 

3_10  =  1 

x      y 

1    \-  1  -1. 

a?     y     z     5 

2a;     32/     2* 

2,3     2_17 

x     y     z     30 

4.   ?_9  =  _2, 

2       3      35 

4     4     3_83( 

9. 

—  —  SB  • 

a;     y 

3a;     2y      6 

a;     2/     z     60 

1-1  =  0. 

•Li-ik 

3        3  _3 

a;     y 

2a;     32/ 

14. 

^""^""i' 

5.   i-3-=-2, 

10. 

^_     _5_  =  _4 

2        3  __1 

x     y 

3a;     22/         3* 

3a;     2y     6' 

5     2     13 

6        3  _19# 

2        3 19 

i  i"'8f 

5a;     2y      5* 

3y     2%          36* 

(3) 

(4) 


LITERAL   SIMULTANEOUS  LINEAR   EQUATIONS        185 
LITERAL  SIMULTANEOUS  LINEAR  EQUATIONS 

Illustration : 

Solve  the  equations :        \  r  ~  '  )J 

{mx  +  ny=d.  (2) 

From  (1),    x  =  l^M.      .From  (2),     x  =  <L=M. 

a  rr 

Therefore,  c-by==d_=lMt 

a  m 

Clearing,  cm  —  bmy  =  ad  —  any. 

Transposing,  any  —  bmy  =  ad  —  cm. 

Therefore,  (an  —  bm)  y  =  ad  —  cm, 

ad  —  cm 


m 


and  y  = 


an  —  bm 


The  labor  of  substituting  a  root  found  for  one  unknown  and  reducing 
the  resulting  expression  for  the  value  of  the  other  unknown,  is  frequently 
as  great  as  that  of  making  a  new  solution  for  the  value  still  undetermined. 
Therefore,  in  practice  it  will  be  well  to  make  a  separate  elimination  for 
each  unknown. 

From(l),     2/  =  ^-=^-       From  (2),     y  =  d  ~™x. 

Therefore,  c_^ax  =  dJZmx. 

b  n 

Clearing,  en  —  anx  =  bd  —  bmx. 

Transposing,  bmx  —  anx  =  bd  —  en. 

Therefore,  (6m  —  an)x  =  bd  —  en, 

and  «=*|izJHL 

bm—  an 

201.  As  in  other  fractional  forms  of  simultaneous  linear 
equations,  solutions  that  do  not  satisfy  both  given  equations 
are  rejected. 

Exercise  65 


1. 

x  +  y  =  m, 
x  —  y  =  n. 

2. 

2x+3y= 

c, 

3x+2y= 

d. 

3.  mx  +  ny  =1, 
nx  —my  =  l. 

4.  mv  +  nw  =  3, 

sv  -f-  tw  =  3. 


186 


SIMULTANEOUS  LINEAR  EQUATIONS 


5.  x  +  y  =  a  +  l,  7.    m(x  +  y)  =  5, 
x  —  y  =  a  —  l.  n(x  —  y)=10. 

6.  x  +  y  =  2a  +  b,  8.   ax  +  cy  =  a  +  2c, 
x-y  =  a  +  2b.  cx+ay  =  a-2c. 

9.   (a  +  l)x  —  (a  —  1)?/ =  4 a, 

(a  +  l)a  +  (a-l)2/  =  2(a2  +  l). 

10.  (m  +  2)a?=(m  — 2)y, 
x  —  n  =  y. 

11.  (a  +  m)»-(a-m)y  =  4am, 
(a  —  m)  &  —  (a  —  ra)  ?/  =  0. 

12.  c»  +  m?/  =  c(c  +  ra)2, 
mx+  cy  =m(c  +  m)2. 


13. 

m     n 

14. 

35      1      y          c 

c  +  1  '  c-l-0' 

c 

15. 

m  .n     1 
cc      2/ 

16. 

2    ,    a       o  . 
—  +  —  =  2  + a, 
ax     2  y 

-  +  -  =  4  +  a2- 
a;     y 

17. 

x              y     _      2 
a  +  1     a  -  1     a2  - 1' 

x     >     y    -     2 

18. 


19. 


x  —  y+  n 

y+x-m=3 

y  —  x—  n 

ax  _  .. 

my 

(a  -f-  m)x      .. 

(ra  ~  a)y 

by 


ax 


2, 


by  +  cz^  2 

a 
cz  +  ax 


21. 


22. 


2. 


ay  ,  bx  _  cy     dx 
6       6  ~9       9 


—  a 


52 

3  ' 


a-1     a+1 


a  —  c 
x-1 
1-c 


^=i, 


+ 


a-1 

y  +  i 
a 


PROBLEMS  —  SIMULTANEOUS  LINEAR  EQUATIONS      187 

PROBLEMS  PRODUCING  SIMULTANEOUS  LINEAR  EQUATIONS  WITH  TWO  OR 
MORE  UNKNOWN  QUANTITIES 

202.  A  problem  may  be  readily  solved  by  means  of  a  state- 
ment involving  two  or  more  unknown  quantities  provided  that 

(1)  there  are  as  many  given  conditions  as  there  are  required 
unknown  numbers,  and 

(2)  there  are  as  many  equations  as  there  are  required  un- 
known numbers. 

Exercise  66 

1.  If  a  certain  number  increased  by  3  is  multiplied  by 
another  number  decreased  by  2,  the  product  is  9  more  than 
that  obtained  when  the  first  number  is  multiplied  by  1  less 
than  the  second  number.  The  sum  of  the  first  number  and 
twice  the  second  number  is  30.     Find  the  numbers. 

Let  x  =  the  first  number, 

and  y  =  the  second  number. 

Then  (a +  3)  (y-2)  =  the  product  of  (the  1st  no. +3)  by  (the  2d  no.-  2). 

Also,        x{y  —  1)  =  the  product  of  (the  1st  no.)  by  (the  2d  no.  —  1). 

Therefore,  from  the  condition, 

(x  +  S)(y-2)-9  =  x(y-l).  (1) 

Also  we  have,  x  +  2  y  =  30.  (2) 

From  (1),  -x  +  3y  =  15.  (3) 

From  (2)  and  (3) ,  y  =  9,  the  second  number. 

x  =  12,  the  first  number. 
Verifying  in  (1) : 

(12  +  3)  (9  -  2)  -  9  =  12(9  -  1). 

(15)  (7) -9  =  12(8). 

105  -  9  =  96. 

96  =  96. 

2.  If  the  larger  of  two  numbers  is  divided  by  the  smaller 
increased  by  5,  the  remainder  is  1  and  the  quotient  3;  but  if 
their  product  decreased  by  43  is  divided  by  1  less  than  the 
larger  number,  the  quotient  is  1  more  than  the  smaller  number. 
Find  the  numbers. 


188  SIMULTANEOUS  LINEAR  EQUATIONS 

3.  If  A  gives  B  $  30,  each  will  have  the  same  amount;  but  if 
B  gives  A  $  30,  the  quotient  obtained  by  dividing  the  number 
of  dollars  A  has  by  the  number  of  dollars  B  has  will  be  -J. 
Find  the  amount  each  has. 

4.  If  1  is  added  to  both  the  numerator  and  the  denominator 
of  a  certain  fraction,  the  result  is  f ;  but  if  2  is  subtracted 
from  the  numerator,  and  2  is  added  to  the  denominator,  the 
fraction  becomes  £.     Find  the  fraction. 

Let  x  =  the  numerator  of  the  fraction, 

y  =  the  denominator  of  the  fraction. 

Then, 

By  the  first  condition, 


By  the  second  condition, 
Solving  (1)  and  (2), 


The  required  fraction  is,  therefore,  -. 


-  =  the  required  fraction. 

y 

x  + 1     8 
y  +  i     9 

(1) 

x-2     1 
«/  +  2     2* 

(2) 

x  =  7,  and  y  =  8. 

5.  A  certain  fraction  becomes  -§-  when  3  is  subtracted  from 
its  numerator  and  4  is  added  to  its  denominator.  The  same 
fraction  is  increased  by  T^-  if  -J-  is  added  to  its  numerator  only. 
Find  the  fraction. 

6.  If  a  certain  fraction  is  divided  by  3,  and  the  result  is 
increased  by  2,  a  new  fraction,  |-£,  is  obtained ;  but  if  the 
original  fraction  is  multiplied  by  2,  and  then  both  numerator 
and  denominator  are  decreased  by  2,  the  result  is  f .  Find  the 
fraction. 

7.  A  certain  number  is  made  up  of  three  digits.  The 
hundreds'  digit  equals  the  sum  of  the  units'  digit  and  the  tens' 
digit;  the  units'  digit  is  3  more  than  the  tens'  digit,  and  the  sum 
of  the  three  digits  is  14.     Find  the  number. 


PROBLEMS  —  SIMULTANEOUS  LINEAR  EQUATIONS       189 

Let  x  =  the  digit  in  the  hundreds'  place, 

y  =  the  digit  in  the  tens'  place, 
z  =  the  digit  in  the  units'  place. 
Then  100  x  +  10  y  +  z  =  the  number. 

From  the  1st  condition,      x  =y  +  z.  (1) 

From  the  2d  condition,        z  —  y         =  3.  (2) 

From  the  3d  condition,        x  +  y  +  z  =  14.  (3) 

Solving  the  system  (1),  (2),  and  (3),  x  =  7,  y=2,  z  =6. 
Therefore,  the  required  number  is  725. 

8.  The  sum  of  the  two  digits  of  a  certain  number  is  13,  and 
if  45  were  added  to  the  number,  the  digits  would  be  reversed. 
Find  the  number. 

9.  If  a  certain  number  is  divided  by  the  difference  of  its  two 
digits,  the  remainder  is  1  and  the  quotient  18 ;  but  if  the 
digits  are  interchanged  and  the  new  number  is  divided  by  the 
sum  of  the  digits,  the  quotient  is  3  and  the  remainder  is  7. 
Find  the  number. 

10.  A  certain  sum  of  money  placed  at  simple  interest 
amounted  to  $  1400  in  3  years,  and  to  $  1500  in  5  years.  What 
was  the  sum  at  interest  and  what  was  the  rate  of  interest  ? 

Let  x  =  the  number  of  dollars  in  the  principal, 

y  ss  the  rate  of  interest. 

The  interest  for  1  year  =  -^-ths  of  the  principal,  =  ^-  dollars. 
100  100 

Therefore,  ^-^  =  the  interest  for  3  years, 

100 

— ^  =  the  interest  for  5  years. 
100  J 

Hence,  3  4.^=  1400.  (1) 

100 

x+^=1500.  (2) 

100  #  Vf 

From  (1) ,   100  x  +  3  xy  =  140000. 


190  SIMULTANEOUS   LINEAR   EQUATIONS 

From  (2),  100  x  +  5xy  =  150000. 

500  x  +  15  xy  =  700000.     (Multiplying  (1)  by  5.) 
300 x  +  15  xy  =  450000.     (Multiplying  (2)  by  3.) 
200  x  =  250000. 
x  =  1250. 
Substituting  in  (1),      y  =  4.  Principal  =  f  1250 1      Resuit 

Rate         =4%      J 

11.  A  sum  of  money  at  simple  interest  amounted  to  $ 336.96 
in  8  months,  and  to  $348.30  in  1  year  and  3  months.  Find 
the  sum  at  interest  and  the  rate  of  interest. 

12.  A  banker  loaned  $15000,  receiving  5%  interest  on  a 
portion  of  the  amount,  and  4  %  on  the  remainder.  The  income 
from  the  5  %  loan  was  $  60  a  year  less  than  that  from  the  4  % 
loan.     What  was  the  sum  in  each  of  the  loans  ? 

13.  Two  loans  aggregating  $  6000  pay  3  %  and  4  %  respec- 
tively. If  the  first  paid  4  %  and  the  second  paid  3  °/0  the  total 
income  from  the  loans  would  be  $  12  more  each  year.  What 
is  the  amount  of  each  loan  ? 

14.  There  are  two  numbers  whose  sum  is  10,  and  if  their 
difference  is  divided  by  their  sum  the  quotient  is  $•.  Find  the 
numbers. 

15.  The  two  digits  of  a  certain  number  are  reversed,  and  the 
quotient  of  the  new  number  divided  by  the  original  number  is 
If,  The  tens'  digit  of  the  original  number  is  4  less  than  the 
units'  digit.     Find  the  number. 

16.  If  the  greater  of  two  numbers  is  divided  by  the  less, 
the  quotient  is  2  and  the  remainder,  1.  If  the  smaller  number 
is  increased  by  20  and  then  divided  by  the  larger  number  de- 
creased by  3,  the  quotient  is  2.     Find  the  numbers. 

17.  Find  two  numbers  such  that  the'  first  shall  exceed  the 
second  by  m ;  and  the  quotient  jof  the  greater  by  the  sum  of  the 
two  shall  be  s. 


PROBLEMS  —  SIMULTANEOUS   LINEAR   EQUATIONS        191 

18.  A  certain  number  of  two  digits  is  9  more  than  four  times 
the  sum  of  the  digits.  If  the  digits  are  reversed,  the  resulting 
number  exceeds  the  original  number  by  18.     Find  the  number. 

19.  A  rectangular  field  has  the  same  area  as  another  field  6 
rods  longer  and  3  rods  less  in  width,  and  also  has  the  same 
area  as  a  third  field  that  is  3  rods  shorter  and  2  rods  wider. 
Find  the  length  and  width  of  the  first  field. 

20.  Two  automobilists  travel  toward  each  other  over  a  dis- 
tance of  60  miles.  A  leaves  at  8  a.m.,  1  hour  before  B  starts 
to  meet  him,  and  they  meet  at  11  a.m.  If  each  had  started  at 
8.30,  they  would  have  met  at  11  also.  Find  the  rate  at  which 
each  traveled. 

21.  Three  pipes  enter  a  tank,  the  first  and  second  together 
being  able  to  fill  the  tank  in  3  hours,  the  second  and  third 
together  in  4  hours,  and  the  first  and  third  together  in  5  hours. 
How  long  would  it  require  for  all  three  running  together  to  fill 
the  tank? 

22.  The  numerator  of  a  certain  fraction  is  the  number  com- 
posed by  reversing  the  digits  of  the  denominator.  If  the 
denominator  is  divided  by  the  numerator,  the  quotient  is  2  and 
the  remainder,  5.  If  the  numerator  is  increased  by  18,  the 
value  of  the  fraction  becomes  1.     Find  the  fraction. 

23.  A  man  seeks  to  purchase  two  different  grades  of  sheep, 
the  whole  to  cost  $210.  If  he  buys  12  of  the  first  grade  and 
13  of  the  second  grade,  he  lacks  $  3  of  the  amount  necessary  to 
buy  them.  If  he  buys  13  of  the  first  grade  and  12  of  the  sec- 
ond grade,  he  still  lacks  $  2  of  the  necessary  amount.  What  is 
the  cost  of  each  grade  per  head  ? 

24.  A  tailor  bought  a  quantity  of  cloth.  If  he  had  bought 
5  yards  more  for  the  same  money,  the  cloth  would  have  cost  f  1 
less  per  yard.  If  he  had  bought  3  yards  less  for  the  same 
money,  the  cost  would  have  been  $1  more  per  yard.  How 
many  yards  did  he  buy  and  at  what  price  per  yard  ? 


192  SIMULTANEOUS   LINEAR  EQUATIONS 

25.    An  automobile  travels  over  a  certain  distance  in  5  hours 
If  it  had  run  5  miles  an  hour  faster,  the  run  would  have  been 
completed  in  1  hour  less  time.     How  far  did  it  run  and  at 
what  rate  ? 


1 

1 


26.  If  the  length  of  a  certain  field  were  increased  by  6  rods, 
and  the  breadth  by  2  rods,  the  area  would  be  increased  by  102 
square  rods.  If  the  length  were  decreased  by  7  rods,  and  the 
breadth  increased  by  4  rods,  the  area  would  be  unchanged. 
Find  the  length  and  breadth  of  the  field. 

27.  A  company  of  men  rent  a  boat,  each  paying  the  same 
amount  toward  the  rental.  If  there  had  been  3  more  men,  each 
would  have  paid  $1  less;  and  if  there  had  been  2  less  men, 
each  would  have  paid  $1  more.  How  many  men  rented  it, 
and  how  much  did  each  pay  ? 

28.  A  boat  runs  12  miles  an  hour  along  a  river  with  the  cur- 
rent. It  takes  three  times  as  long  to  go  a  certain  distance 
against  the  current  as  it  does  with  it.  Find  the  rate  of  the 
current  and  the  rate  of  the  boat  in  still  water. 


29.  144  voters  attend  a  meeting  and  a  certain  measure  is 
passed.  If  the  number  voting  for  it  had  been  ^  as  large,  and 
the  number  voting  against  it  had  been  f  as  large,  the  vote 
would  have  been  a  tie.  How  many  voted  for  and  how  many 
against  the  measure  ? 

30.  The  total  weight  of  three  men  is  510  pounds.  The  first 
and  third  together  weigh  340  pounds,  and  the  second  and  third 

together  350  pounds,  and  the  first  and 
second  330  pounds.  Find  the  weight 
of  each  of  the  three. 

31.  The  sum  of  the  three  angles  of 
a  triangle  is  180°.  The  angle  at  A  is 
£  of  the  angle  at  B\  and  the  angle 

at  C  is  f  the  sum  of  the  angles  at  A  and  B.     Find  the  number 

of  degrees  in  each  angle. 


DISCUSSION  OF  A  PROBLEM  193 

THE  DISCUSSION  OF  A  PROBLEM 

203.  Since  a  problem  is  solved  by  means  of  equations  based 
upon  given  conditions,  it  follows  that  a  true  solution  can  result 
from  possible  conditions  only.     Briefly, 

(a)  Impossible  conditions  result  in  no  solution  ;  or, 

(&)  An  impossible  answer  indicates  impossible  conditions. 

Some  important  cases  are  considered  under  the  following 
heads : 

(a)  A  Negative  Result  may  indicate  an  Impossible  Problem 

1.  If  a  dealer  doubles  the  number  of  horses  he  owns  and 
also  buys  9  additional  head,  he  will  then  have  -J-  the  number 
he  might  have  possessed  by  adding  20  head  to  5  times  the 
original  number.     Find  the  number  he  originally  possessed. 

Let  x  =  the  original  number  of  horses. 

2  x  +  9  =  the  number  under  the  first  condition  named. 
x  +  20  =  the  number  under  the  second  condition. 

Then  2x  +  9  =  6-*±™. 

O 

6x  +  27  =  5z  +  20. 
6x-  5z  =  20-27. 

x=-7. 

And  the  negative  result  indicates  an  impossible  problem,  due  to  a  fault 
in  the  statement  of  the  conditions. 

If  the  statement  had  been  "and  then  sells  9  head"  instead  of  "and 
also  buys  9  additional  head  "  the  problem  would  have  been  possible. 

For  2x-9  =  6x  +  20, 

3 

6x-  27  =  5z  +  20. 

x  =  47.     Result. 

(b)  A  Statement  in  a  Problem  may  be  Reversed 

2.  A  certain  man  42  years  of  age  has  a  son  18  years  old. 
How  many  years  ago  was  the  father  twice  as  old  as  the  son  ? 

SOM.    EL.    ALG. 13 


194  SIMULTANEOUS   LINEAR   EQUATIONS 


Let  x  =  the  number  of  years  since  the  father  was  twice  as    ! 

old  as  the  son. 
Then       18  —  x  =  the  son's  age  x  years  ago. 

42  —  x  =  the  father's  age  x  years  ago. 
From  the  condition, 

42  -x  =  2(18  -x). 
42  -  x  =  36  -  2  x. 
2sc-x  =  36-42. 

x  —  —  6.     Result. 

The  result  indicates  that  6  years  have  still  to  elapse  before  the  condi- 
tion named  will  hold.  (That  is,  in  6  years  the  father  will  be  48  and  the 
son  24  years  of  age.)  If  the  problem  were  changed  to  read  "  In  how 
many  years  will  the  father  be  twice  as  old  as  the  son  ?  "  the  solution 
would  be  possible. 

(c)  A  Fractional  Result  may  indicate  an  Impossible  Problem 

3.  If  the  number  of  boys  in  a  certain  schoolroom  is  de- 
creased by  5,  there  will  be  left  2  more  than  one  third  the 
original  number.     How  many  boys  were  there  at  first  ? 


Let  x  =  the  number  of  boys  at  first. 

From  the  given  condition, 


X 

-5 

=  |+2. 

Sx- 

-15 

=  a  +  6, 

Sx 

—  X 

=  6  +  15. 

2x 

=  21. 

X 

=  10*. 

Clearly,  the  fractional  result  indicates  an  impossible  condition. 

(d)  Possible  Discussions  of  Given  Values  and  their  Relations 
to  each  Other 

A  general  problem  admits  of  discussion  by  assuming  differ- 
ent relations  between  given  values. 


DISCUSSION  OF  A  PROBLEM  195 

4.  Two  trains,  an  express  and  a  mail,  pass  along  the  same 
railroad  in  the  same  direction,  the  express  train  traveling  m 
miles  per  hour,  and  the  mail  n  miles  per  hour.  At  12  o'clock 
the  mail  is  k  miles  ahead  of  the  express.  In  how  many  hours 
will  the  two  trains  be  together  ? 

We  may  assume  that  they  are  together  x  hours  after  12  o'clock,  the 
express  having  traveled  mx  miles,  and  the  mail  nx  miles.  But,  by  the  con- 
ditions, the  express  has  traveled  k  miles  more  than  the  mail  at  12  o'clock ; 

hence, 

mx  —  nx  =  k. 

k 


m  —  n 
Discussion : 

1.  Suppose  m  is  greater  than  n. 
The  value  of  x  will  be  positive. 

The  express  will  overtake  the  mail  after  12  o'clock. 

2.  Suppose  m  less  than  n. 
The  value  of  x  will  be  negative. 

(For,  if  n  is  greater  than  w,  the  rate  of  the  mail  was  the  faster  rate.) 
The  trains  were  together  before  12  o'clock. 

This  assumption,  therefore,  is  impossible,  for  we  are  going  contrary  to 
the  condition  that  the  trains  are  to  be  together  after  12  o'clock. 

3.  Suppose  m  equal  to  n. 

The  value  of  x  will  become  _. 
0 

If  the  rate  m  equals  the  rate  n,  the  trains  have  not  been  together,  are 

moving  at  the  same  constant  distance  apart,  and  will  never  be  together. 

In  this  case,  therefore,  the  supposition  has  led  to  the  impossible  con- 

k 
dition  denoted  by  the  symbol  -. 

This  symbol  is  usually  denoted  by  oo,  and  is  read  "  infinity." 

4.  Suppose  m  equal  to  w,  and  k  equal  to  0. 

If  k  equals  0,  the  mail  and  the  express  started  together ;  and  since  m 
equals  n,  the  trains  have  been,  and  will  continue  to  be,  together. 
Expressing  this  final  condition  in  symbols,  we  have, 

x  =  — - —  =  -  =  any  finite  number. 
m  —  n     0  • 


196  SIMULTANEOUS  LINEAR  EQUATIONS 

That  is,  there  is  an  infinitely  great  number  of  points  at  which  the  two 
trains  are  together.  In  this  case,  therefore,  -  is  the  symbol  of  inde- 
terminate value. 

In  general,  therefore : 

From  (2),  A  negative  result  indicates  an  error  in  statement. 
From  (3),  A  result  x  =  —  =  co  indicates  no  possible  solution. 

From  (4),  A  result  x  =  —  indicates  an  infinitely  great  number  of  solu- 
tions. 


Exercise  67 

Discuss  the  following  problems  and  interpret  the  solution 
for  each : 


1.  A  is  12  years  old,  and  B  is  17  years  old.  In  how  many 
years  will  B  be  twice  as  old  as  A  ? 

2.  The  total  number  of  boys  and  girls  in  a  certain  school 
is  32,  and  four  times  the  number  of  boys  plus  twice  the  num- 
ber of  girls  equals  95.  How  many  boys  and  how  many  girls 
are  there  in  the  school  ? 

3.  A  and  B  can  together  paint  a  sign  in  8  hours,  and  B 
alone  can  paint  the  same  sign  in  5  hours.  How  many  hours 
will  A  require  if  working  alone  ? 

4.  A  boy  has  45  coins,  the  value  being  in  all  78  cents. 
A  portion  of  the  number  consists  of  nickels,  and  the  remain- 
der of  cents.     How  many  are  there  of  each  kind  ? 

5.  A  group  of  b  boys  bought  a  boat,  agreeing  to  pay  d  dollars 
each,  but  /  of  the  boys  failed  to  pay  their  shares,  and  each 
remaining  boy  had  to  pay  e  dollars  more  than  he  had  agreed. 
Find  the  cost  of  the  boat  in  terms  of  b,  e,  and  / 

Ans.     (6~/> 6e. 
/ 


CHAPTER   XVII 


THE   GRAPHICAL  REPRESENTATION  OF  LINEAR 
EQUATIONS 

THE  GRAPH  OF  A  POINT 

In  the  figure  two  straight  lines,  XX'  and  YY',  intersect  at 
right  angles  at  the  point  0,  the  origin.     These  lines,  XX'  and 

YY'j  are  axes  of  reference. 
From  O  measure  on  OX 
the  distance,  OA  =  a.  From 
0  measure  on  OF  the  dis- 
tance, OB  —  b.  Through  A 
draw  a  line  parallel  to  0  Y, 
and  through  B  draw  a  line 
parallel  to  OX.  These  lines 
intersect  at  P.  And  P  is 
the  graph  of  a  point  plotted 
by  means  of  the  measure- 
ments on  OX  and  OY. 

204.    OA     and     OB,   or 

their  equals,  a  and  6,  are 
the  rectangular  coordinates  of  point  P. 

205.  A  coordinate  parallel  to  the  XX'  axis   is  called  an 
abscissa. 

A   coordinate    parallel  to  the  YYf    axis   is   called  an   or- 
dinate. 

206.  Abscissas  measured  to  the  right  of  0  are  positive,  to 
the  left  of  0,  negative. 

197 


Y 

B 

P 

/> 

x 

a 

X 

0 

jA 

Y 

198 


GRAPHS   OF  LINEAR   EQUATIONS 


Y 

P 

M 

X 

X 

O 

T 

N 

Y 

Ordinates  measured  upward  from  0  are  positive,  downward 
from  0,  negative.     In  the   accompanying  diagram  it  will  be 

seen  that 

The  abscissa  of  P  is  +  3  ;  the 
ordidate,  +  4. 

The  abscissa  of  M  is  —  4 ;  the 
ordinate,  +  3. 

The  abscissa  of  N  is  —  4 ;  the 
ordinate,  —  5. 

The  abscissa  of  T  is  +  2  ;  the 
ordinate,  —  3. 

207.  It  will  be  seen  that 
the  axes  of  reference  di- 
vide the  plane  of  the  axes 
into  four  parts  or  quad- 
rants, and  these  quadrants  are  named  as  indicated,  I,  II,  III 
and  IV.  From  the  principle  governing  the  signs  (Art.  206),  , 
we  observe  that  in 

Quadrant  I :   abscissa  +  ;  ordi- 
nate +  • 

Quadrant  II :  abscissa  — ;  ordi- 
nate +  . 

Quadrant  III :  abscissa  —  ;  ordi- 
nate — . 
Quadrant    IV  :     abscissa  +  ; 

ordinate  — . 

In  naming  the  coordi- 
nates of  a  point,  the  ab- 
scissa is  always  named  first, 
and  the  ordinate  last. 

In  the  illustrations  of  this  chapter  the  scale  of  the  graphs  is 
so  chosen  that  one  unit  of  division  on  the  axes  corresponds  to 
one  numerical  unit  in  a  solution.  In  later  chapters  the  student 
will  easily  apply  other  scales  as  occasion  requires. 


Y 

It 

X"      _ x 

— -g 

II                 IE 

7^ 

THE   GRAPH  OF  A  POINT 


199 


In  the  diagram  let  the 
student  name  the  location 
of  each  of  the  several  points, 
making  a  table  in  which 
the  position  of  each  is  re- 
corded. 


4=  (2,  4), 

**< 

), 

B=  (-5,1), 

0=( 

)i 

C=(-3,-5), 

R={ 

), 

#=(7,  -7), 

K={ 

), 

fc=(           ), 

£  =  ( 

)• 

Y 

H 

A 

K 

B 

X 

X 

G 

O 

L 

F 

E 

C 

D 

Y 

Exercise  68 

On  properly  ruled  paper  plot  the  following  points : 
I  1.   (2,5).  6.    (-3,  4).  11.    (4,0). 

2.  (3,4).  7.    (5,  -2).  12.    (0,4). 

3.  (7,1).  8.    (3,3).  13.    (-5,0). 

4.  (1,7).  9.    (-2,-1).  14.    (0,-5). 


15.    (0,  0). 
5)? 


5.    (-2,5).  10.    (-4,-7). 

16.  In  what  quadrant  does  (-  3, 5)  lie  ?  (2,  -4)?  (-3, 
(4,  7)? 

17.  In  what  quadrant  does  (5.5,. 7.5)  lie  ?  (-  4.8,  -  9.75)  ? 
(-  5.4,  8.7)  ?     (-  14.75,  -  6.1)  ? 

18.  Using  two  divisions  on  your  paper  for  one  given  unit  of 
measure,  plot  the  points ;  (4,3),  (2f  5)f  (-2,3),  (3,-2), 
(0,  4),  and  (7,  0). 

19.  Using  one  division  on  your  paper  for  two  given  units  of 
measure,  plot  the  points;  (4,  6),  (10,  18),  (-8,  14),  (20,-16), 
(-18,-18),  (0,-24),  (10,15),  (-17,-21),  (-3,25), 
(25,  -3),  and  (-19,  -12). 


200 


GRAPHS   OF  LINEAR   EQUATIONS 


THE  GRAPH  OF  A  LINEAR  EQUATION  IN  TWO  UNKNOWN  NUMBERS 

208.    A  variable  is  a  number  that,  during  the  same  discus- 
sion, may  have   an   indefi- 


X 

A 

J>r- 

\B 

s 

n 

5s 

Sn 

S?     _ 

Ss  - 

X"~ 

m SEL 

-x 

5 —    ~SC 

s 

V 

S^ 

^ 

_       B 

7^ 

nitely     great 
values. 


number     of 


If     2/=0,  *=4,  P=(4,0). 

y=l,  x=3,  P1=(3,  1). 

y=2,  a>=2,  P2=(2,2). 

2/=3,  a=l,  P3=(l,3). 


209.    A    constant    is    a 

number  that,  during  the 
same  discussion,  has  one 
and  only  one  value. 

If,  in  the  given  linear 
equation,  jc-f  y  =  4,  we  as- 
sume a  succession  of  differ- 
ent values  for  y,  the  corre- 
sponding values  of  x  are  as 
follows: 

2/=4,  0=0,      P4=(0,4). 
y=5,  a=-l,P5=(-l,5). 
y=6,  x=-2,P6=(-2,6). 
etc.,  indefinitely. 


Kegarding  the  respective  pairs  of  values  as  the  coordinates 
of  points  which  we  may  designate  by  P,  Plt  P2,  etc.,  we  plot 
these  points  and  draw  through  them  the  line  AB.  Therefore, 
the  line  AB  in  the  figure  is  the  graph  of  the  linear  equation, 
x  +  y  =  4. 

An  indefinite  number  of  points  might  have  been  obtained  by 
assuming  fractional  values  for  y,  and  finding  the  corresponding 
values  of  x. 

Exercise  69 

Plot  the  graphs  of  the  linear  equations : 

1.  x  +  5y  =  7.  3.    x  +  3y  =  ±.  5.    x+y  =  0. 

2.  3x  +  y  =  2.  4.    2x  +  7y  =  3.  6.   #-4?/  =  0. 


LINEAR   EQUATIONS   IN   TWO  UNKNOWNS 


201 


A  Shorter  Method  for   Obtaining    the   Graph  op   a 
Linear  Equation  in  Two  Variables 

210.  It  can  be  proved  that  the  graph  of  every  linear  equa- 
tion in  two  variables  is  a  straight  line.  (This  fact  justifies  the 
use  of  the  word  "linear"  in  naming  so-called  equations  of 
the  first  degree.)  Now  a  straight  line  is  determined  by  two 
points,  hence  the  graph  of  a  linear  equation  should  be  deter- 
mined by  the  location  of  any  two  points  that  lie  in  its  graph. 

The  two  points  most  easily  determined  are  those  where  the 
graph  cuts  the  axes.     Therefore, 

(1)  Find  the  point  where  the  graph  cuts  OX  by  placing  y  =  0, 
and  calculating  x. 

(2)  Find  the  point  where  the  graph  cuts  OFby -placing  x  —  0, 
and  calculating  y. 

Illustration: 
Plot  the  graph  of 

Sx- 
If  t/  =  0,  x  =  2. 
Plot  Pi  (2,0). 

If  x  =  0,  y  =  -  3. 
PlotP2  (0-3). 

Join  PiP2. 

AB  is  the  required  graph  of 
3cc-2y=6. 


2y  =  6. 


Y 

Q-/k ™~" 

mmmm 

b ZZZ—Z 


Exceptions  to  the  Shorter  Method 
(a)  A  Linear  Equation  whose  Graph  passes  through  the  Origin. 
211.  If  a  given  linear  equation  has  the  form  of  ax  =  by,  it  is 
evident  that  when  x  =  0,  y  =  0  also.  That  is,  a  graph  of  such 
an  equation  passes  through  the  origin.  Therefore,  to  plot  the 
graph  of  an  equation  in  this  form,  at  least  one  point  not  on 
either  axis  of  reference  must  be  determined. 


202 


GRAPHS   OF    LINEAR    EQUATIONS 


(b)  A  Linear  Equation  whose  Graph  is  Parallel  to  Either  Axis. 

212.    If  a  given  linear  equation  is  in  form  of  #=7,  it  is  evident 

that  the  value  of  a?  is  a  constant.     That  is,  the  abscissa  of  every 

point  in  the  graph  is  7.     Therefore,  the  graph  of  this  equation 

is  a  straight  line  parallel  to  the  axis,  YY'  and  7  units  to  the 

right  of  it. 

Exercise  70 

Plot  the  graphs  of  the  following : 

1.  x+2y  =  5.        5.   a?  =  4.  9.  Sx  —  5y  =  —  3. 

2.  y  =  3x  +  4:.         6.   y  +  5  =  0.  10.  5x  +  3y  =  0. 

3.  x  =  2y  +  3.        7.   3x-2y  =  -10.  11.  12 a; -17 2/ =15. 

4.  x  +  ±y  =  5.        8.   y  =  7  x.  12.  4?/ =  16  —  18  a. 


THE   GRAPHS    OF    SIMULTANEOUS    LINEAR    EQUATIONS   IN   TWO   UN- 
KNOWN  NUMBERS 

(a)  Independent  Equations 

In  the  figure,  the  line  AB 
is  the  graph  of  the  linear 
equation,  x  +  y  =  5.  The 
line  CD  is  the  graph  of  the 
linear  equation,  4  a?—  3y =6. 
It  will  be  seen  that  the 
graphs  intersect  at  the 
point  P,  (3,  2).  That  is, 
the  point,  (3,  2),  is  common 
to  both  graphs.  Solving  the 
given  linear  equations, 
x  +  y  =  5  and  3  x  —  4  y  =  1, 
we  obtain  as  a  result,  x  =  3, 


-si 


* 


^ 


y  ss  2.     And,  clearly,  the  values  obtained  for  a;  and  2/  are  the 
coordinates  of  the  intersection  of  the  graphs. 

213.  The  coordinates  of  the  point  of  intersection  of  the  graphs 
of  two  simultaneous  linear  equations  form  a  solution  of  the  two 
equations  represented  by  the  graphs. 


SIMULTANEOUS  LINEAR   EQUATIONS 


203 


Exercise  71 


Solve  the  following  simultaneous  linear  equations  and  verify 
the  principle  of  Art.  213  by  plotting  their  graphs : 


1.  5x  —  3y  =  l, 
Sx  +  5y  =  21. 

2.  5a -3?/ =  36, 
7x—  5y  =  56. 


3.  8a +  3^  =  12, 
12  a  +  5y  =  16. 

4.  4x  +  6y  =  -3, 
2</  +  a  =  0. 


(ft)  Inconsistent  Equations 


Given  two  equations 
Multiply  (1)  by  2, 


2a  +  3i/  =  8, 
4*  +  6y=  -25. 

4se  +  6y  =  16. 


(1) 
(2) 


The  equations  are  therefore  inconsistent,  for  it  is  impossible 
to  find  any  values  of  x  and  y  that  will  satisfy  both  equations. 
Plotting  the  graph  of 
2x  +  Sy  =  8  (^4jB),  and  the 
graph  of  £x  +  6y  =  —  25 
(CD),  we  obtain  two  parallel 
lines,  and  as  parallel  lines 
never  meet,  we  find  no  point 
common  to  the  graphs. 
Thus  the  graphical  repre- 
sentation of  the  linear 
equations  in  this  case  shows 
with  great  clearness  the 
real  meaning  of  inconsist- 
ent equations,  and  serves 
to    emphasize    more    than 

ever  the  need  of  independent  equations  if  a  solution  is  to  be 
possible. 


1 

A 

ss 

^-C                        ^    "" 

5Cjv 5*-J-     .X 

*S"~  o"      •  S~ 

X                         ^B 

^^ 

s 

^^ 

D 

Y^ 

204 


GRAPHS  OF  LINEAR  EQUATIONS 


(c)  Indeterminate  Equations 


Y 

B 

X 

1 

X 

0 

A 

Y 

simultaneous  equations  by  means  of  the  graph 


Given  two  equations 
2x  —  3y  =  10,         (1) 

4a -6  ?/ =  20.         (2) 
Multiply  (1)  by  2, 
4  x  -  6  y  =  20. 

That  is,  the  first  equa- 
tion is  put  in  the  form 
of  the  second,  and  their 
graphs  are  found  to  coin- 
cide. Here,  also,  we  fail 
to  find  independent  equa- 
tions; and  we  are  again 
assisted  to  a  clearer  con- 
ception   of   indeterminate 


Exercise  72 

Apply  the  principle  of  graphical  representation  to  the  follow- 
ing, and  determine  the  pairs  of  independent,  inconsistent,  and 
indeterminate  equations. 


1.  hx  +  2y  =  l, 
5a  +  2  y  =  -S. 

2.  3x-±y  =  7, 
9x  —  12y  =  0. 

3.  3  x  +  ly  =  —  4, 
4a-3y-7  =  0. 

4.  4  x  -f-  y  —  9, 
3#-22/  =  4. 

5.  2x  —  5y  =3, 
6x  =  l&  +  15y. 


6.  4  x  —  5  y  =  3, 
8a>-10y  =  6. 

7.  2a>  —  3y  =  l, 
16  oj  -  24  y  =  8. 

8.  a?+%  =  19, 
lly-3a>  =  19. 

9.  4#-32/  =  8, 
16-8a;  =  -6y. 

10.    3a;  —  5*s%, 

9z  =  15  +  18y. 


CHAPTER   XVIII 

INVOLUTION  AND  EVOLUTION 
INVOLUTION 

214.  Involution  is  the  operation  of  raising  a  given  expression 
to  any  required  power.  All  cases  of  involution  are  multiplica- 
tions, the  factors  in  each  case  being  equal.  In  all  elementary 
work  the  exponents  of  the  powers  are  positive  and  integral. 

THE  GENERAL  PRINCIPLES  FOR  INVOLUTION 

(a)  The  Power  of  a  Power 

When  m  and  n  are  both  positive  integers : 

By  Art.  61,  am  =  a  x  a  x  a  ...  to  m  factors. 

Therefore,  (am)n  =  (a  x  a  x  a  ...  to  m  factors)  (a  x  a  x  a  ...  to 

m  factors)  ...  to  n  groups  of  factors, 
=  a  x  a  x  a  ...  to  mn  factors, 
as  amn.  The  Third  Index  Law. 

Hence,  to  obtain  any  required  integral  power  of  a  given 
integral  power  : 

215.  Multiply  the  exponent  of  the  given  power  by  the  exponent 
of  the  required  power. 

(b)  The  Power  of  a  Product 

When  n  is  a  positive  integer : 

(a&)n  =  ab  x  ab  x  ab  •••  to  n  factors, 
=  (a  x  a  x  a  •••  to  n  factors)  (b  x  b  x  b  •••  to  n  factors), 
=  (a«)(6»), 

=  anbn.  The  Fourth  Index  Law. 

205 


206  INVOLUTION  AND   EVOLUTION 

Hence,  to  obtain  any  required  power  of  a  product : 

216.  Multiply  the  factors  of  the  required  product,  first  raising 
each  factor  to  the  required  power. 

(c)  The  Power  of  a  Fraction 

When  n  is  a  positive  integer : 

(a\n     a  /.  a  „  a  -    . 

[  -     =7xTx7-ton  factors, 
\bj       b     b     o 

—  (a  x  a  x  a  '"  to  n  ^actors) 
(b  x  b  X  b  •••  to  n  factors) 

~b» 
Hence,  to  obtain  any  required  power  of  a  fraction : 

217.  Divide  the  required  power  of  the  numerator  by  the  same  I 
required  power  of  the  denominator. 

These   laws   are   general  for  unlimited  repetitions   of  the 
powers  or  of  the  factors. 

Thus :        C(on)m]p  =  amnp,  etc.,  (abc  •••  w)m  =  ambmcm  •••  nm,  etc. 

THE  SIGNS  OF  POWERS 

218.  All  even  powers  of  any  quantity,  positive  or  negative,  are 
positive. 

Thus,  (+a)2  =  +  a2.    (+  a)4  =  +  a4.     (-a)2=+a2.     (-  a)4  =  +  a4. 

219.  j£B  odd  powers  of  any  quantity,  positive  or  negative,  have 
the  same  sign  as  the  given  quantity. 

Thus,  (+ a)8  =  +  a3.    (+a)6=+a5.     (-a)8=-a8.     (-a)5=~a5, 

(a)   The  Involution  of  Monomials 
Illustration : 

1.    I  -1*LY '..(«*^'.-«^  ■    Result. 
V      3m»y<y  (3«V)'  27WV2 


INVOLUTION  OF  BINOMIALS  207 

In  general,  to  raise  a  monomial  to  a  required  power : 

220.  Determine  the  sign  of  the  power.  Raise  each  factor  of 
the  given  quantity  to  the  required  power,  the  numerical  factors  by 
actual  multiplication,  and  the  literal  factors  by  multiplying  each 
exponent  by  the  exponent  of  the  required  power. 

Oral  Drill 

Give  orally  the  value  of: 

\3  mnj  \       xy  )' 

2.    (3xy.  8<    (a*x\\  (     4aVy 

W       14-  rwj- 
3-(-2^-     9-(fSJ-     njlm 

ii  (2x^w 

5.    (2mrc2)5.  "    lgTO«J 


2^2/V* 

16.    _(^_5^Y 
V      3cdJ 


17.    -(|m3nY)6. 
6.    (3aV)4.  _V^M  18-    -(-faWrc)4. 


-  -&>' 


(b)   The  Involution  of  Binomials 

The  process  of  raising  a  binomial  to  a  required  power,  or,  the 
expansion  of  a  binomial,  is  best  shown  by  comparative  illustra- 
tions. 


By  actual  multiplication : 

(a  +  6)2  =  a2  +  2  ab  +  62. 

(a  +  6)3  =  a3  +  3  a26  +  3  a62  +  63. 

(a  +  6)4  =  a4  +  4  at6  +  6  aib2  +  4  a?>3  +  64. 

(a  +  6)5  =  a5  +  5  a46  +  10  efib*  +  10  a2^  +  5  ab*  +  &5. 


208  INVOLUTION  AND  EVOLUTION 

* 

221.   From  a  study  of  the  form  of  each  product  we  note : 

1.  The  number  of  terms  in  the  expansion  exceeds  by  1  the  ex- 
ponent of  the  binomial. 

2.  The  exponent  of  a  in  the  first  term  is  the  same  as  the  expo- 
nent of  the  power  to  which  the  binomial  is  raised,  and  it  decreases 
by  1  in  each  succeeding  term. 

3.  b  first  appears  in  the  second  term  with  an  exponent  1,  and 
this  exponent  increases  by  1  in  each  succeeding  term  until  it  is  the 
same  as  the  exponent  of  the  binomial. 

4.  TJie  coefficient  of  the  first  term  is  1,  and  of  the  second  term 
the  same  as  the  exponent  of  the  binomial. 

5.  The  coefficient  of  each  succeeding  term  is  obtained  from  the 
term  preceding  it,  by  multiplying  the  coefficient  of  that  term  by 
the  exponent  of  a,  and  dividing  the  product  by  the  exponent  of  b 
increased  by  1. 

6.  If  the  second  term  of  the  given  binomial  is  positive,  the  sign 
of  every  term  of  the  expansion  will  be  positive;  and  if  the  sign  of  | 
the  second  term  of  the  given  binomial  is  negative,  the  signs  of  the 
terms  of  the  expansion  will  be  alternately  positive  and  negative. 

Illustrations : 

1.  Expand  (a  — #)4. 

For  the  a-factor :  a4        a3  a2  a 

x-f  actor:                         x            x2  Xs      x* 

coefficients:             4            6  4 

signs :  -  + -  + 

Combining,  a4  —  4  a8x  +  6  a2x2  —  4  ax*  +  x*.    Result. 

2.  Expand  (3 a -|Y. 

By  observing  the  formation  of  the  expansion  of  (a  —  &)6,  we  write : 
(a  -  h)'°  =  a6  -  5  a46  +  10 a862  -  10 a263  +  5 a&4  -  b&. 
(3a-^-(3a)5_5(3a)4^+10(3a)3^2_10(3a)2(|y 


EVOLUTION  209 

Simplifying, 

2  4  16         32 

A  polynomial  is  raised  to  a  power  by  the  same  process,  if 
the  terms  are  so  grouped  as  to  express  the  polynomial  in  the 
form  of  a  binomial. 

3.   Expand  (1-a  +  ar2)3. 

(1  -  X  +  Z2)3  =[(1  -  x)  +  *2]3 

=  (1  -  X)3  +  3(1  -  X)*X*  +  3(1  _  s)(z2)2  +(>2)8 

=  (1  _  Sx  +  3x2  -  x*  +  3x2  -  6x8  +  3x*  +  3x4  -  3x6  +  36) 
=  (1-  3x  +  6x2  -  7x8  +  6x*  -  3x6  +  x6).    Result. 


Exercise  73 
Expand : 

1.   (m  +  w)3.  8.   (3a-l)4. 


13. 


(*-i)r- 


2.  (m  +  ny.  9.   (ax-2yy. 

3.  (m-*)6-  i0.   (c2-^)5.  14-  (^  +  ^  +  l)2- 

4.  (c  +  3)4.  /       3.N3  15.  (m2-2m  +  3)2. 

5.  (a-2)5.  n'   (C  +  2y*  16-  C^  +  c  +  l)3. 

6.  (3a-2y)\*  /        _2V  17«  (rf-x  +  lf- 

7.  (2c2-3)3.  I  37  18.  (2aJ-aj2  +  2)3. 


EVOLUTION 

222.  Evolution  is  the  process  of  finding  a  required  root  of  a 
given  expression. 

223.  The  symbol  for  a  required  or  expressed  root  is  the 
radical  sign,  ^/. 

Thus  :  Va  =  the  square  root  of  a  ;  y/a  =  the  cube  root  of  a. ;  etc. 

224.  The  number  written  in  the  radical  sign  and  indicating 
the  root  required  is  the  index  of  the  radical. 

SOM.    EL.    ALG.  —  14 


210  INVOLUTION  AND   EVOLUTION 

Thus  :  3  is  the  index  of  the  indicated  cube  root  above.  It  will  be  noted 
that  in  the  case  of  a  square  root  the  index  is  not  usually  written,  the 
absence  of  an  index  being  an  "  understood  "  square  root. 

The  vinculum  is  commonly  used  to  inclose  the  expression  affected  by  a 
radical. 

'  225.   From  the  definition  of  root  (Art.  60)  it  is  clear  that 
y/a  means  "  Required  :  One  of  the  two  equal  factors  of  a." 
v/&  means  "  Required  :  One  of  the  three  equal  factors  of  &." 
y/c  means  "  Required  :  One  of  the  four  equal  factors  of  c." 
{Vx  means  "  Required  :  One  of  the  n  equal  factors  of  x,"  etc. 

THE  GENERAL  PRINCIPLES  OF  EVOLUTION 

In  the  discussion  of  these  principles  both  m  and  n  are  posi- 
tive and  integral  numbers. 

(a)  The  Root  of  a  Power 

By  Art.  215,  (am)n  =  amn. 

Therefore,  by  definition  (Art.  60), 

am  is  the  nth  root  of  amn. 
For  am  is  one  of  the  n  equal  factors  of  (am)n. 

That  is,  am  =  {Var™.  The  Fifth  Index  Law. 

The  conclusion  is  a  direct  result  of  a  division  of  the  exponent  of  the 
given  quantity  by  the  index  of  the  required  root.     Or, 

mn 

y/a^n-  —  a/"  =  am. 
Hence  : 

226.  Any  required  root  of  a  power  is  obtained  by  dividing  the 
exponent  of  the  power  by  the  index  of  the  required  root. 

(5)  The  Root  of  a  Root 
By  Art.  226,  (mKVa)mn  =  a. 

Extracting  the  nth.  root,      ("\/a)m  =  Va. 
Extracting  the  with  root,       (w-v/a)  =  Zjtya. 


GENERAL  PRINCIPLES   OF  EVOLUTION  211 

Hence : 

227.    The  mnth  root  of  an  expression  is  equal  to  the  mth  root 
of  the  nth  root  of  the  expression. 

(c)  The  Root  of  a  Product 


By  Art.  226, 

({Vab)n  =  ab. 

Therefore, 

(Va  x  {Vb)n  =  (Va)n  x  ({Vb)n. 

Or, 

(y/aby=(Vax  y/iy. 

Hence, 

{Vab={Vax  Vb. 

Therefore : 

228.   Any  required  root  of  a  product  of  two  or  more  factors  is 
equal  to  the  product  of  the  like  roots  of  the  factors. 


w 

The  Root  of  a  Fraction 

By  Art.  217, 

/a\n_an 
\b)       b» 

Therefore, 

>&»     b 

Hence : 

229.  Any  required  root  of  a  fraction  is  obtained  by  finding  the 
like  roots  of  its  numerator  and  denominator. 

THE  SIGNS  OF  ROOTS 

(a)  Positive  Even  Powers 

By  Art.  218,  (+  a)(+  a)  =  +  a2  and  (-  a)(-  a)  =  +  a2 

Therefore,  V+a2  =  +  a  or  —  a. 

Hence : 

230.  Every  positive  number  has  two  square  roots  whose  abso- 
lute value  is  the  same,  but  whose  signs  are  opposite  in  kind. 

The  double  sign,  ± ,  is  used  to  indicate  two  roots.     Thus,  Va2  =  ±a. 
The  sign  ±  is  read,  "  plus  or  minus." 


212  INVOLUTION   AND   EVOLUTION 

(b)  Negative  Even  Powers 
Since         —  a2  =  (+  a)(—  a),  we  have  a  product  of  unequal  factors. 

Hence : 

231.    It  is  impossible  to  obtain  an  even  root  of  a  negative 
number. 

(c)  Positive  and  Negative  Odd  Powers 

ByArt.219,  (+a)(  +  a)(+a)  =  -f  a3.    Also,  (_a)(_a)(-a)  =  -a8, 
etc. 


Therefore,       V+  a3  =  -f  a,   V— a8  =  —  a,  etc. 
Hence : 


232.  !Z7ie  odd  roots  of  a  positive  quantity  are  positive,  and  the 
odd  roots  of  a  negative  quantity  are  negative;  or,  briefly,  the  odd  I 
roots  of  a  quantity  bear  the  same  sign  as  the  given  quantity. 

THE  EVOLUTION  OF  MONOMIALS 

233.  Illustrations : 

1.  Eequired  the  cube  root  of  8  a%3c9. 

The  root  is  odd  ;  the  given  quantity,  positive  ;  the  sign  of  the  result,  +. 
Dividing  each  exponent  by  the  index  of  the  root, 

\/S  a*b*c*  =  \/2*cfib*<P  =  2  a?bc8.    Eesult. 

2.  Eequired  the  fifth  root  of  -  243  xl0y15. 

The  index  is  odd ;  the  given  quantity,  negative ;  the  sign  of  the  result,  — . 
Dividing  each  exponent  by  the  index  of  the  root, 


V-243x1(y6  =  V  -  36£c102/15  =  -  3  x2y*.    Result. 
3.   Eequired  the  fourth  root  of  16  aWc12. 

The  root  is  even;  the  given  quantity,  positive;  the  result  bears  the 
double  sign. 

Hence  :         ^16  a*b*c12  =  \/2*a8&*c12  =  ±  2  a26c3.    Result. 


SQUARE  ROOT  OF  POLYNOMIALS        213 

4.   Required  the  square  root  of  1587600. 

A  root  of  a  large  number  may  frequently  be  obtained  from  its  prime 
factors. 

Hence,  V 1587600  =  V2*  •  3* .  52  .  72  =  ±  (22 .  32 .  5  .  7)  =  ±  (4  .  9  .  5  •  7) 
=  ±  1260.    Result. 

In  the  consideration  of  numerical  quantity  we  shall  consider  only  the 
positive  roots  in  our  results. 


Oral  Drill 

Find  the  value  of : 

1.  V121  a4c6.  10.  -v/-243m10z30.      18.      4/625^ 

2.  V64m47iy.         11.  vSSW.  16zl6 

3.  VWtftf.  12.  ^-1024m15n20.  19.    Jj/EZ. 

4.  ^-27  aft/3.         13.  -\/64a%18.  64<f 

5.  ^-343  (W.       14.  ^729  aW°.  20'    <\F 

6.  \/16m&/iM.  i5.  ^/i28c¥"4. 


243  m15 


21        3/125^" 

7.    ^81  <Wy.        16.    ^256m<V«.  '    V"27V  * 

9.    -^-64aW.  N64w6  \1024ic30 

23.    V1296.  24.    ^3375. 

THE  SQUARE  ROOT  OF  POLYNOMIALS 

234.  If  a  binomial,  (a  +  6),  is  squared,  we  obtain  (a2  +  2  ab 
+  ft2).  We  have  in  the  following  process  a  method  for  extract- 
ing the  square  root,  (a  +  6),  of  the  given  square,  (a2  +  2  a&  +  62). 


a2  +  2  a&  +  62  |  a  +  6  square  root.        The  first  term  of  the  root, 

a? a,  is  the  square  root  of  the 

«6  +  6 


b 


: 

given  expression,  the  remainder,  (2  ab  +  62),  results.     Dividing  the  first 
term  of  this  remainder,  2  ab,  by  twice  the  part  of  the  root  already  found, 


+  2  ab  +  62  first  term  of  the  given  ex- 

-f  2  a&  +  62  pression,  a2. 

Subtracting  a2    from   the 


214 


INVOLUTION  AND   EVOLUTION 


2  a,  the  quotient  is  b,  the  second  term  of  the  root.  This  second  term,  b,  is 
added  to  the  trial  divisor,  2  a  ;  and  the  sum,  2a-\-b,  is  multiplied  by  b. 
The  result,  (2a+&)6,  =  2«H  b'2,  and  completes  the  process.  It  will  be 
seen  that  much  of  the  work  depends  directly  upon  the  trial  divisor,  2  a. 
The  reason  for  this  prominence  of  the  trial  divisor  will  be  seen  from  the 
following 


a* 

ab 

ab 

V 

Graphical  Representation  of  a  Square  Root 

235.   In  the  accompanying  figure  we  have  a  graphical  repre- 
sentation of  a  square  constructed  upon  a  given  line,  a  +  b.     If 
the  square  whose  area  is  a2  is  subtracted 
from  the  whole  area,  there  remain  the  areas 

ab  +  ab+b2=2ab  +  b2. 

(Note  that  this  corresponds  with  the  sub- 
traction above.)  Now  the  length  of  the 
side  of  the  square  removed  being  a,  we  are 
to  provide  for  a  remaining  area  built  upon 
two  sides  of  that  square,  or  a  +  a  =  2  a  i 
(the  trial  divisor).     Now 


ajb 


ab 


& 


Area 
Length 


=  Width.      Hence, 


2ab 
2a 


b. 


That  is,  b  is  the  width  of  the  remaining 
area  whose  length  is  known.  Now,  because 
of  its  position  in  the  original  square,  there 
still  remains  unprovided  for  the  square  62, 
whose  side  is  b  in  length.  Hence  the  length  of  the  total  area 
necessary  to  complete  the  square  is  2  a  -\-b.  (This  explains 
the  addition  of  the  second  term  of  the  root  to  the  trial  divisor 
above.)  Multiplying  our  known  length  by  the  width  ascer- 
tained by  division,  we  have,  as  above, 

(2a  +  b)b  =  2ab  +  b2, 

which  area  completes  the  square  required. 


SQUARE   ROOT  OF   POLYNOMIALS 


215 


236.  By  the  principle  of  Art.  234,  we  obtain  the  square  root 
of  any  polynomial,  the  trial  divisor  at  any  point  being  in  every 
case  twice  the  part  of  the  root  already  found. 

Illustrations : 

1.   Extract  the  square  root  of  xA  -f-  6  Xs  -f-  19  x2  -f-  30  x  +  25. 

A  polynomial  must  be  arranged  in  order  if  its  root  is  to  be  found 
without  difficulty. 

x*46x3  +  19x2  +  30x+25|x2+3x+5 

x4 Result. 

First  Trial  Divisor,    2  (x2)  =  2  x2  +  6  x3 + 1 9  x2 

First  Completion,       (2  x2+ 3 x) (3  x)  —  \  +6  x3+  9 x2 


Second  Trial  Divisor,  2  (x2 + 3  x)  =  2  x'2 + 6  x 
Second  Completion,    (2x2+6  x+5)(  +  5)  = 


+  10x2+30x+25 
+  10x2+30x+25 


2.   Extract  the  square  root  of  1  —  2  a  to  4  terms. 

The  approximate  square  root  of  expressions  not  in  themselves  perfect 
squares  is  obtained  by  the  principles  of  Art.  191. 


2a 


+ 


»(1)  =  2 

(2-q)(-q) 


2a 

2«  +  a2 


2(1 -a)  =2-2  a 

t-"-s(-a 


-a2 


a2+a8  +  «L 


('-'-*-f)(-f)- 


4 

2       4 


Result. 


In  extracting  the  square  root  of  fractional  expressions  care  must  be 
iken  that  the  expression  is  properly  arranged.     The  descending  powers 
of  a  letter  occurring  in  both  numerator  and  denominator  of  a  fraction  are 
written  thus  : 

o8  +  0a  +  0  +  i+I  +  I+  1  +  I . 


3.   Extract  the  square  root  of  —  + 11  -\ h  - ■  H — - 

x2  x       az       a 


216  INVOLUTION   AND   EVOLUTION 

In  descending  powers  of  a : 


*  +  y  +  u  +  £*  +  *«.+  B4.«.    ReSUit. 
x2      x  a      a2\x  a 


x2 


2^=  — 
\xj       X 

(¥+3)(+3)  = 

a; 
a; 

2(«+?\     ^  +  6 

\X          )         X 

\  x              a/  \     a/ 

+  2+  —  +  -! 
a       a2 

+  2+  —  4~ 
a      a2 

Exercise  74 

Extract  the  square  root  of : 

1.  a4  +  4<e3  +  6£c2+4a;+l. 

2.  a4-4a34-10ff2-12a;4-9. 

3.  a4-6a3  +  12a  +  4  +  5a2. 

4.  a6  +  4^-2a;4-l6a?  +  ^  +  12a;  +  4. 

5.  13m2-30m  +  4ra4+20ra3+9. 

6.  4  a;4  +  4  y?y  +  9  ic2?/2  +  4  ai/3  -f-  4  y4. 

7.  9a4-12a3&  +  34a2&2-20a&3  +  25&4. 

8.  25c4-20c3d-4cd3+14c2d2  +  d4. 

9.  4-12»  +  17aJ»-32a?  +  34a!4-20aJ,  +  25a* 

10.  16a^-16afy-28a^  +  30a^-20a;y+29afy4  + 

11.  ^  +  3a;  +  9. 
4 

12.  a;4  +  2jK3+2^4-a;4-i. 


25  V6. 


13. 


13  a2 


2+    4 


3a +9. 


SQUARE  BOOT  OF  ARITHMETICAL  NUMBERS        217 


14.    !l2  +  2a  +  3+2_c+c°. 
&       c  a       ar 

4  a;4      4  a?     7  a?     x     1 
'9  9         9       3     4* 

9  a;4     6  a?  .13  a?     x       1 
'     25        5    +   10       2+16* 

4  a2     2  a     a?     19     g 
'    9ar>     3a;     a2     12     a' 

9  9 ■'■  T  4  T  9   ^16^  4 

Find  three  terms  of  the  square  root  of : 

19.  1+9  x.  21.   ar2-}-^.  23.    9^  —  1. 

20.  a?-S.  22.   4  0^-5  0;.  24.    36 -12a;. 


THE  SQUARE  ROOT  OF  ARITHMETICAL  NUMBERS 

237.  Since  an  arithmetical  square  integer  is  the  result  of 
the  multiplication  of  some  integer  by  itself,  we  are  assisted  in 
obtaining  arithmetical  square  roots  by  noting  a  certain  relation 
that  exists  between  such  numbers  and  their  squares. 

I2  =           11  A  number  of  one  place  has  not  more  than  two 

92  =         81 J  places  in  its  square. 

102  =       100 1  A  number  of  two  places  has  not  more  than  four 

992  =     9801 J  places  in  its  square. 

1002  =    10000 1  A  number  of  three  places  has  not  more  than  six 

9992  =  998001 J  places  in  its  square. 

Conversely,  therefore : 

If  an  integral  square  number  has  two  figures,  its  square  root  has  one 
figure. 

If  an  integral  square  number  has  four  figures,  its  square  root  has  two 
figures. 

If  an  integral  square  number  has  six  figures,  its  square  root  has  three 
figures. 


a 

+  2  ab  +  b'2\a  +  b                            1 
2                                          302  = 

a  +  b 
1296 180  +  6     Result. 
900 
396 
396 

__ 

+  2  ab  +  62         2  (30)  =  60 
+  2ab  +  b2         (60  +  6)  (6)  = 

218  INVOLUTION  AND   EVOLUTION 

Hence : 

238.  Separate  any  integral  square  number  into  groups  of  two 
figures  each,  and  the  number  of  groups  obtained  is  the  same  as 
the  number  of  figures  in  its  square  root. 

Illustrations : 

1.   Find  the  square  root  of  1296. 

Beginning  at  the  decimal  point, 
separate  into  periods  of  two  figures 
Parallel  Algebraic  Process        each. 


(*)*  = 

2(a)  =  2  a 
(2  a +  6)  (6) 

In  the  square  root  of  1296  :  The  greatest  square  in  1296  is  900. 

The  square  root  of  900  is  30. 

The  trial  divisor  is  2  (30)  =  60. 

The  second  term  of  the  root  is  (396  -f-  60  =6). 

For  the  completion,  (60  +  6)  (6)  =  396. 
The  process  is  repeated  in  the  same  order  if  the  given  integer  is  of  a 
higher  order. 

2.   Eind  the  square  root  of  541,696. 

The  following  process  is  given  in  a  form  commonly  used  in  practice. 

Separating  into  periods  of  two  figures  each :  ,  •   •   • 

-n,     i        ,.  541696   /So 

Explanation  : 

The  greatest  square  contained  in  the  first  period 
(54)  is  49.  The  square  root  of  49  is  7.  7  is,  there- 
fore, the  first  figure  of  the  root.  Subtracting  49 
from  54,  and  bringing  down  the  two  figures  of  the 
next  period,  we  have  516,  the  remainder.  Annex- 
ing a  0  to  the  first  figure  of  the  root,  7,  our  trial  divisor  is  2  (70)  =  140. 
Dividing  516  by  140,  we  obtain  3,  the  second  figure  of  the  root. 
(140  +  3)  3  =  429,  which  product  is  subtracted  from  516.  With  the 
remainder  (87),  we  bring  down  the  last  two  figures  (96),  and  the  new 
remainder  is  8796.  Annexing  a  0  to  the  figures  of  the  root  already  found, 
our  trial  divisor  is  2  (730)  =  1460.    Dividing  8796  by  1460,  we  obtain  6, 


0 

541696 

140  +  3 

49 

1516 

3 

[429 

1460  +  6 

8796 

6 

8796 

SQUARE   ROOT   OF   ARITHMETICAL   NUMBERS         219 


the  third  figure  of  the  root.      (1460  +  6)6=8796,  which  product,  sub- 
tracted from  8796,  gives  a  remainder  of  0,  and  the  square  is  completed. 

The  addition  of  the  0  to  the  figures  of  the  root  already  obtained  gives 
a  trial  divisor  of  the  same  order  as  the  remainder,  or  of  the  next  lower 
order.    The  process  gives  fewer  figures  and  less  likelihood  of  error. 

If  a  given  square  number  has  decimal  places,  we  point  off  by 
beginning  at  the  decimal  point,  first  separating  the  whole 
number  as  before,  finally  separating  the  decimal  into  periods 
from  left  to  right.  Ciphers  may  be  annexed,  if  necessary,  to 
complete  any  period. 

If  a  given  number  is  not  a  perfect  square,  its  approximate 
square  root  can  be  found  to  any  desired  number  of  places. 

The  square  root  of  a  common  fraction  is  best  found  by  chang- 
ing the  fraction  to  a  decimal  and  extracting  the  square  root  of 
the  decimal  to  the  required  number  of  places. 

Illustrations : 

1.  Find  the  square  root  of  2.  Find,  to  three  decimal 
19920.4996.  places,  Vl2f. 


20  +  4 
4 

280  +  1 
1 
2820  +  1 
1 
28220  +  4 
4 


19920.4996| 
1 

99 

96 


320 
281 


3949 

2821 


141.14  f  =  .375.  12|  =  12.375. 

Result.  The  required  three  decimal  places 

necessitate  six  decimal  figures  in  the 
square.  Hence,  with  three  ciphers 
annexed,  we  have  to  obtain  the 
square  root  of 

12.37500013.517+.-. 
9  Result. 


112896 
112896 


The  decimal  point  in  the  result 
is  located  easily  by  noting  between 
which  periods  of  the  given  example 
the  given  decimal  point  lies.  In  the 
example  above  there  are  two  periods 
to  the  right  of  the  decimal  point 
given ;  therefore,  there  will  be  two 
decimal  places  in  the  root  obtained. 


60  +  5 
5 

337 
325 

700  +  1 
1 

1250 
701 

7020  +  7 

7 

54900 
49189 

220  INVOLUTION  AND   EVOLUTION 

Exercise  75 

Find  the  square  root  of : 

1.  9216.    5.  186624.    9.  .717409.    13.  .00002209. 

2.  67081.   6.  4202500.   10.  9617.7249.   14.  .0001752976. 

3.  32761.   7.  49.434961.  11.  44994.8944.  15.  .009409. 

4.  182329.   8.  9486.76.   12.  .00119716.   16.  .0000879844. 
Find,  to  three  decimal  places,  the  square  root  of : 

17.  3.      19.  7.      21.  .5.      23.  f      25.  .037. 

18.  5.  20.  10.  22.  .05.  24.  f.  26.  .0037. 
Find,  to  three  decimal  places,  the  value  of : 

27.  5  +  3  V'2~.  29.  2V3-V5T  31.  Ve  +  VK 

28.  V7+-VI0.  30.  3V7-7V3.  32.  ^VS-Vf. 

33.  How  many  rods  in  the  side  of  a  square  field  whose  area 
is  2,722,500  square  feet? 

34.  Simplify  and  extract  the  square  root  of 

^(10^  +  13)-2(2a4  +  3  +  7^)a;  +  (arJ  +  l)  (a4-  a2  +  1). 

35.  Simplify  and  extract  the  square  root  of 

(x2  -  2x  -  3)  (x2  -  x  -  6)  (x2  +  Sx  +  2). 

36.  Show  that  the  required  square  root  in  the  preceding 
example  can  be  readily  obtained  by  factoring  and  inspection. 

37.  If  a,  b,  and  c  are  the  sides  of  a  right  triangle,  and  a  lies 
opposite  the  right  angle,  we  may  prove  by  geometry  that 
a2  =  b2  +  c2.  Find  the  length  of  a  in  a  right  triangle  in  which 
b  and  c  are  210  feet  and  350  respectively. 

38.  Applying  the  principle  given  in  the  preceding  example, 
find,  to  three  decimal  places,  the  distance  from  the  "  home  1 
plate  to  the  second  base ;  the  four  base  lines  of  a  regulation 
baseball  diamond  forming  a  square  90  feet  on  each  side. 


CHAPTER   XIX 
THEORY   OF   EXPONENTS 

239.  The  Index  Laws  for  Positive  Integral  Values  of  m  and 
n  have  been  established : 

(1)  For  Multiplication:  am  x  an  =  am+n  (61) 

(2)  For  Division  (m  greater  than  n)  :  am  ■*  an  =  am~n  (79) 

(3)  For  a  Power  of  a  Power :  (am)n  =  amn  (215) 

(4)  For  a  Power  of  a  Product :  (ab)n  =  anbn  (216) 

(5)  For  a  Root  of  a  Power :  Va"*™  =  am  (2~6) 

240.  The  extension  of  the  practice  of  algebra  requires  that 
these  laws  be  also  extended  to  include  values  of  m  and  n  other 
than  positive  integral  values  only;  and  the  purpose  of  this 
chapter  is  to  so  extend  those  laws.     We  shall,  therefore : 

1.  Assume  that  the  first  index  law  (am  x  an  =  am+n)  is  true  for 
all  values  of  m  and  n. 

2.  Define  the  meaning  of  the  new  forms  that  result  under  this 
assumption,  m  or  n,  or  both,  being  negative  or  fractional. 

3.  Show  that  the  laws  already  established  still  hold  true  with 
our  new  and  broader  values  for  m  and  n. 

THE  ZERO  EXPONENT 

If  m  and  n  may  have  any  values,  let  n  =  0. 

Then,  am  x  a°  =  aTO+°  (61) 

=  am 

dm 
Dividing  by  am,  a0  =  — 

That  is,  a°  =  1. 

Hence,  we  define  a0  as  equal  to  1.    Or : 

221 


222  THEORY   OF   EXPONENTS 

241.    Any  quantity  with  the  exponent  0  equals  1. 

Illustrations : 

1.   s°  =  1.  2.    (wm)°  =  1.        3.     3  m°  =  3-1 

4.   4aP  +  (2y)°  =  4.1  +  l=5. 


Oral  Drill 

Simplify  orally : 


1.  a°x.         4.   4a°6°c.       7.   5  a? -5.        10.   3a0  +  30  +  a°60. 

2.  2aa°.      5.   3a°  +  l.    8.   2a°+3a°.    11.   2a°  +  (a°-l). 

3.  3aV-    6-   4a°-3.    9.   4a;0- 52/°.    12.   4?x  +  (a  +  3 a?)0. 

THE  NEGATIVE  EXPONENT 

If  m  and  w  may  have  any  values,  let  n  be  less  than  0  and 
equal  to  —  m. 

(61) 
(241) 


Then, 

am  x 

a-7"  =  amr~m 

lence, 

am  x 

a-m  =  1. 

Dividing  by  a~m, 

1 

Dividing  by  am, 

1 
a_m  =  —  • 
a™ 

Hence,  we  define  ar™  as  1  divided  by  am. 

From  this  definition  we  obtain  an  important  principle  of  con- 
stant use  in  practice : 

242.  Any  factor  of  the  numerator  of  a  fraction  may  be  trans- 
ferred to  the  denominator,  or  any  factor  of  the  denominator  may 
be  transferred  to  the  numerator,  if  the  sign  of  the  exponent  of  the 
transferred  factor  is  changed. 

Illustrations : 

1    <,&-*-«  2    ar^s-i-JL.  <*    2-lmx-*_3mn 


FRACTIONAL   FORM   OF   THE    EXPONENT  228 

Oral  Drill 

Transfer    to    denominators     all    factors    having     negative 
exponents : 

1.  ax~2.         4.   2a2~b~2.  7.     crtc"3.  10.   2  ar^ar1. 

2.  crf-V"3.     5.  2-1ar2b-s.        8.   x~ly-h-\         11.   5  ary3*4- 

3.  2ab~2.       6.    Stf-Vr1-       9-   S-'a-^c-1.       12.   iHar^Mr^. 

Read  the  following  without  denominators : 

a2                  cd  Sa  c2  1 

13.    -•  15. 17.    ~-       19.    -^r-0.       21. 


b2                 mn              a?y2           '   ar2b~2  2~2x~2f 

xyz               ax          m    5  x        AA-     2  a  ab 

14.    -2-.        16.    — •        18.    — •        20.    -rt.        22.  53  2 

m                 zJ                a  l               m6n  t  —3  "era 

Read  the  following  with  all  exponents  positive : 
23.    5a-1&c.  27.    =    ;    ,-  31. 


2ar>"1  2-1m2w-12 

24.  12  a-1^.  28.    5 5-5-5*  32.  r — rr— j. 

25.  -5«*V.  29.    — 53.  33.     3^. 

26.  3a-Vd  30.    |S^C  34. 


3-V^z  -  3  m-1^-^  y-1^2 

THE  FRACTIONAL  FORM  OF  THE  EXPONENT 

■ 

If  the  expression  a2  can  be  shown  to  conform'  to  the  first 
index  law,  we  may  find  a  definition  for  the  fractional  form  of 
exponents.     By  the  first  index  law, 

(a*)2  =  at  x  J  =  a?+%  =  a. 

Hence  the  meaning  of  a%  is  established,  and  the  exponent  in 
this  form  still  agrees  with  the  fundamental  index  law. 
That  is: 


224  THEORY  OF  EXPONENTS 

a?  =  Va  is  one  of  the  two  equal  factors  of  a. 
Similarly,       (a^)3  =  eft  x  eft  X  eft  =  cft+*+$  =  eft  =  a2. 

2 

That  is,  a3  is  one  of  the  three  equal  factors  of  a2. 

In  like  manner,  a?  =  Va  =  one  of  the  three  equal  factors  of  a. 

3  4  . 

a*  =  va3  =  three  of  the  four  equal  factors  of  a. 
Therefore : 

243.  In  the  fractional  form  of  an  exponent  we  may  define  the 
denominator  as  indicating  a  required  root,  and  the  numerator 
as  indicating  a  required  power. 

Illustration : 

Vs2    =  8»    =  22  =  4.     And  4  contains  two  of  the  three  equal  factors  of  8. 
\/8P  =  81^=38=27.    And  27  contains  three  of  the  four  equal  factors  of  81. 
In  general : 

m        m       m  m    m    m 

an  •  an  •  an  •••  to  n  factors  —  an  n  n  •••  to  n  terms 

mn 

=  am. 

m 

That  is,  an  is  an  expression  whose  nth  power  is  am. 

m 

And        an  is  the  nth  root  of  am- 

m 

Or,  as  above,    an  is  one  of  the  equal  factors  of  am.  % 

It  is  understood  that  while  a*  must  equal  either  +  Va  or  —  Va,  we 

i  r- 

consider  the  positive  value  only  ;  and  a2  is  denned  as  -f  V a,  the  principal 

square  root  of  a. 

In  future  operations  we  may  apply  the  definition  of  Art.  243 
to  expressions  given  in  radical  forms,  observing  that 

244.  The  index  of  a  radical  may  be  made  the  denominator  of 
an  exponent  in  the  fractional  form,  the  given  exponent  of  the 
power  of  the  quantity  becoming  the  numerator  of  the  fractional 
form. 


FRACTIONAL  FORM   OF  THE   EXPONENT  225 

Illustrations : 

1.    Vrn*  =  wA  2.    Vtf  =  x\  3.    y/b^  =  b^. 

And  in  the  converse  operation  of  the  principle  of  Art.  244  : 

4.   **=v^.  5.    a$=Va?.  6.   c^=«v^ 

In  the  reduction  of  numerical  forms  : 

7.    v/l66  =  (\/l6)5=(2)5=32.       8.    v^8*  =  (\/=8)6  =  (-2*)  =  -32. 

It  is  to  be  noted  that  the  root  is  extracted  first. 

Oral  Drill 

Express  with  radical  signs : 

1.  eft.  4.   cfix?.  7.   y*z\  10.   x$y%z$. 

2.  x\  5.   m^/*.  8.    cW.  11.   rftp^q*. 

3.  c*.  6.   «&*.  9.   e^mi  12.    ir&rfix?. 
Express  with  exponents  in  fractional  form : 

13.  Va?.        16.    Vz\         19.    y/a  •  Vc3.  22.  V4~a*  •  V^"8. 

14.  Vm5.      17.    -v^S*.        20.    V^-^c3.  23.  ^^-V3674. 

15.  Va7.        18.    <^.        21.    ^/m^.-^/n3.  24.  VlScSJ* •  \/8?. 
Give  the  numerical  value  of: 

25.  4*  28.    64*.  31.      4*.  34.    81*.         37.   32^. 

26.  9i  29.      4*.  32.      8*.  35.    25*.         38.    64*. 

27.  27*.  30.      8*.  33.    27*.  36.    49*.         39.    128* 
Having  established   a  meaning  for  the  new  forms  of  ex- 

m 

ponents,  a0,  a-1,  and  an,  we  must  show  that  the  index  laws  hold 
true  for  these  new  forms;  thus  fulfilling  the  third  and  final 
clause  of  our  agreement  in  Art.  239. 

SOM.    EL.    ALG.  15 


226  THEORY  OF  EXPONENTS 


Proof  of  the  Index  Laws  for  Negative,  Fractional,  and 
Negative  and  Fractional  Values  of  m  and  n 

245.   In  the  following  proofs,  m  and  n  are  rational  integers 
or  rational  fractions. 

The  Law  amxan  =  am+n. 

1.  When  m  and  n  are  negative  and  integral. 

a~m  x  a~n  —  —  x  —  = =  a-*-*. 

am     <xn     aw+n 

2.  When  m  and  n  are  positive  and  fractional. 

Let  m  =  ■£  and  n  =  -,  »,  q,  r,  and  s  being  positive  and  integral. 
q  s 

P  r  ps  qr  ps+qr 

a«  x  a*  =  a«*  x  a«*  =  VaP*  x  Va«r  =  Va**  x  a*r  -  tyap+Hr  =  a    «8 

=  a«    • 

3.  When  m  and  n  are  negative  and  fractional. 

Let  wi  =  —  £  and  n  —  —  I,  p,  g,  r,  and  s  being  positive  and  integral. 

-p-     --    i     i      i       -M-9- 

p  r  p+r 

a?     a*     a?    » 

Let  the  student  discuss  this  law  when  m  is  a  positive  and  n  a  negative 
fraction. 

TheLaw(aw)n  =  amn. 

1.  When  m  and  n  are  negative  and  integral. 

Let  m  =  —  p  and  w  =  —  q,  p  and  a  being  positive  and  integral. 
(am)n  =  (a~p)~9  =(—)     =  (ap)q=  a**  —  a(-*X-«)  =  aw.  > 

2.  When  w  is  jpo^iYwe  and  fractional. 

Let  w  =^,  p  and  g  being  positive  and  integral. 
Q 

P  mp  m  p 

(am)n  =  (awl)<*  =  V(am)p  =  Vam»  =  a  «  —  a  '«, 


APPLICATIONS   OF  PKINCIPLES   OF  EXPONENTS  227 

3.   When  n  is  negative  and  fractional. 

Let  n  =  —  ■?,  p  and  q  being  positive  and  integral. 
0. 


=  — =a    «  =a  v   «y 


(am)n=(am)    *  = 

(am) *        « 


The  Law  (a6)w  =  anbn. 

1.  When  n  is  negative  and  it 
Let  n  —  —  p,  p  being  positive  and  integral. 

(ab)m  =  (db)-p  =  — —  =  — ?—  =  a-vb-P. 
(aby     aPbP 

2.  When  n  is  positive  and  fractional. 

Let  n  =  -,  p  and  <?  being  positive  and  integral. 

p         p  p. 

(ab)m  =  (aft)  «  =  ty(ab)p  =  -^a*>&p  =  i'a*'  •  VbP  =  a*  6« 

3.  When  w  is  negative  and  fractional. 

Let  n  =  —  -,  p  and  q  being  positive  and  integral. 

-E         i  i  _£    _£. 

(«&)"»=(«&)    «  = ;  =  "T7  =  a    ?6    q 

(ab)i     aib* 

Let  the  student  discuss  this  law  (1)  when  m  and  n  are  both  positive 
and  fractional,  and  (2)  when  m  and  n  are  both  negative  and  fractional. 


APPLICATIONS  OF  THE  PRINCIPLES  OF  EXPONENTS 

(a)  Simple  Forms  involving  Integral  Exponents 

246.  In  processes  with  exponents  no  particular  order  of 
method  can  be  said  to  apply  generally.  Experience  with  dif- 
ferent types  will  familiarize  the  student  with  those  steps  that 
ordinarily  produce  the  clearest  and  best  solutions.  As  a  rule, 
results  are  considered  in  their  simplest  form  when  written 
with  positive  exponents. 


228  THEORY   OF  EXPONENTS 

247.    Illustrations : 
1.   Simplify  x~A  •  x5  •  x~7. 

x-*  •  sc5  -  x-7  =  z-4+6-7  =  a;-«=  — .    Result. 

x6 


2.  Simplify  m3n~2  •  m~2n~5 

m8n-2  •  m_2n-6  =  wi3_ 

3.  Simplify  ^^  -  ^Vl 

ar1         (ma;-1)  ~J 


m8n-2  •  m-2n-5  =  wi3-%-2"5  =  mn~7  =  S .    Result. 

ML  7 


^m-gft4        (a"2)3    _a2m~3y>cg-6 


X"1 

(jnix-1)-2      x-im-W 

=  a2-6m~ 

5+2^4+1-2 

=  a^w-ix3 

- 

-_^!_.   Result. 
a4m 

Simplify : 

Exercise  76 

1.   a5  x  a-3. 

5.   m2  •  m~3  •  m5. 

9.    x-3  +  x-2. 

2.   c~3  x  c7. 

6.   x*  •  ic°  •  or4. 

10.    m~4  -r-  m~7. 

3.   ra"2  X  m3. 

7.   a-V-1^-8^ 

11.   a~sm~2 -r- a2m~3. 

4.   a;-4  •  x3. 

8.   c^d-cM"4. 

12.   ar1^"5  +  x~5y~l. 

13. 

C" 

2dV3 
'd-'z5 

x®y~lm3n~9 
m5n~7x°y-8 

14. 

a 
a 

~2b- 

~2b 

"2^°.               16. 

a?m~hr3x~5 
a~5m~3n5x7 

17.   (a"3)2. 

22.    (»-^l|)"1. 

27.    (a°  +  ay)-3. 

18.  (c-3)-2. 

19.  (a)"5. 

23.  (a-V2)2. 

24.  (a-1^-2)3. 

28.    (mV)°x(3a;0)2. 

20.  (a"1)5. 

21.  (a-lb)\ 

25.  (a2a°a-3)-2. 

26.  (c-1^  •  c0^-1)- 

,  >°-m'- 

APPLICATIONS  OF  PRINCIPLES  OF  EXPONENTS      229 

(6)   Types  involving  the  Fractional  Form 

248.  In  the  following  illustrations  attention  is  called  to 
each  important  feature  of  the  process,  and  the  order  of  the 
principles  that  is  emphasized  in  each  is  such  as  will,  under 
similar  conditions,  produce  the  best  form  of  solution. 

Illustrations : 

1.  Simplify  (a^V1)"2. 

(asrtr1)"2  =  <r4&^2  =  h^~-  •     Result. 

Note  (1)  that  the  first  step  is  the  application  of  the  law  (am)n  =  amw, 
and  (2)  that  the  result  is  given  with  all  exponents  positive. 

2.  Simplify  (c2V03. 

(c2  \/<Fi)8  =  (c2  •  c~*)8  =  (cfy  =  A    Result. 
Note  that  the  law  am  x  an  =  am+n  is  first  applied  so  as  to  unite  c-f actors. 

3.  Simplify  {^(V^)-|}~* 


Note  that  the  reduction  is  accomplished  outward. 


Result. 


L   Simplify  ( 


9Vm- 


s\-f 


>25Vm/ 

/9  v^F2\-f  __  p^-f  _  /     9     \"i  =  /25  mM=  25?  ro*  _  125  m^ 
V25Vm/         V26mi/  \25mV  V     9     /    "       9*  27 

Result. 
Note  that  in  the  third  step  inverting  the  fraction  changes  the  sign  of 
the  exponent  of  the  fraction.    In  general,  ( -\  ~*  =  225  =  £  =  (k\ x. 


230  THEORY  OF  EXPONENTS 

Vm f  Vn~i  _     jm  Vn"4     _  ["w^n""*  _     /mn~2  T 


=  [mJrT*  -T-  Vm%-8]6 

b=  (wi*n~^  -h  m«_i)6 

=  (m^VM)* 

=  (m^)6 

=  m2.     Result. 

Exercise  77 

Simplify : 

1.    (8a-2a"1)i 

6.    (^Txyt+i-l/Slx7)-1. 

if 


3/ =-  7.    JaY(*X\Z 

2.  \(aV-3Vaary.  Af    *  V2T7         2/ 

3.  [m^m-^m-?)2]6.  g    ^  c~ V    §  Jm*  y/tfc  _ 

r a~Wm        m*a16Vc 

4.  [Va-ici(c-W)^]"8. 


r  3/ \ 1-2  9.     \ r-  X = 

5.    L^ax    -Vax-U    '  c*d-2  cy/d 


10 


A   c~Wx         4/I6  n    Jc^n* 


n     As  a  ~s/x2+  27  xi  Va-<aQ-* 
(9a?*-+-4at) 


12. 


[yV  <  Vy3 


L^2   ^    y^. 

13.    |Jm3a;-1^/m2a,-2%/^v 


APPLICATIONS   OF  PRINCIPLES  OF  EXPONENTS      231 

(c)  Types  involving  Numerical  Quantities  Only 

249.  The   exponents   of   a   numerical   expression  must  be 
made  positive  before  reduction  is  attempted. 

Illustration : 

Simplify  9*  +  27"$  +  5°  -  8"*  + 1. 

9t  +  27~8  +  50_8-f  +  i  =  33  +  JL+1__L+1 

27*  8$ 

=  27  +  |  +  1  -  |  +  1 
=  28§£.    Result. 

Exercise  78 

Simplify : 

l-   (f)-3  +  (f)-2-(t)3-  3.  4  a?+ (4  *)•+!*•. 

2.  (J)-'- 8* +  (!)-«.  4.  -^s-s^+c^)-1. 

6.    16^  +  16«0-16°-(^-)-2. 

•[8-^-'!][©,+<^>(r+(r} 

(rf)  Types  involving  Literal  Exponents 

250.  Illustration : 

m  m_i  2m— 1    n* 

Simplify  [a»      -a"     -^a^7"]   . 

p    m  m  2m—    -i«x         r-    m+n  m— n  2m    l-.nx 

\_an+  •  a»~  ■+■  a    «~  J     =  [_a  «     •  a~^*~-r-  a-**- _ 

Cm+n      m— n_2m— 1-ji 
a~n~       «  «~J 


a*.    Result. 


232                               THEORY   OF  EXPONENTS 

Exercise  79 

Simplify : 

1.  a2m+n  •  am+n-!-a?m+n.  7     f  *+i  .     1  \^ 

2.  (m*)x+1  +  mx+2.  K          ^J 

3.  («*)•*■  +  (<**•).  8.    ^_2n+2x8r 

4.  (m*-")'  (my~*y  -*-  m_(ae+2)y. 


16" 


5.      (03&    2C6)ax. 


9-  [^'(^t^)^} 


ax+2y         ay+z         ax+3a  4~ T~ 


(e)   Miscellaneous  Processes   involving    the    Principles   op 

Exponents 

251.   Illustrations : 

1.   Multiply  a  -  3  J  -  2  a$  by  2  -  a*  +  3  a~i 
g  —  3  gf  —  2  g^ 


2- 

-  a  "i  +  3  g~$ 

2a- 

-  6  g$  -  4  g$ 

-    gt  +  3gi  +  2 

+  3  a\  _  9  _ 

-6a 

-1 

2  a  -  7  of  +  2  rf  -  7  -  6  g~£    Result. 
2.   Divide  9  a*  -  3  a*  + 1  +  7  <T*  -  6  a~*  by  3  +  cf*  -  2  cf*. 

9g*  -  3 a*  +  1  +  7  g"?  -  6 a-*  (3-f  g-i-2g~£ 

9  gi  +  3  g£  -  6 3  g£  -  2  g£  +  3.    Result. 

-  6  gi  +  7  +  7  o~i 

-6gi-2+4g"? 

+  9  +  3g_?-6g"i 
+  9  +  3  g~i  -  6  g"i 


APPLICATIONS   OF  PRINCIPLES   OF   EXPONENTS      233 

3.  Simplify  qq  +  q^  +  q-^i-q). 


a (1  +  a)-1  +  <rl(l  ~  «)  _  j+g 


a      .  1  —  a 


a(l  +  a)_ 1 


I -a      2a2-l      2  a2  - 1 


Result. 


1  -t-a        a 


a  (1  +  a) 


4.  Expand  (a2 -2  a"2)4. 

(a2-2a-2)4  =  (a2)4-4(a2)3(2a-2)  +  6(a2)2(2a-2)2_4(a2)(;2a-2)3+(2a-2>* 
=  a8  -  8  a4  +  24  -  32  a~*  +  16  <r8.     Result. 

5.  Extract  the  square  root  of 

--^-2^  +  ^  +  17  +  12^5  +  4^ 


tf*-2aT*-3-2a*.    Result. 


-  4  art  -  2  af  £ 

-  4  aT?  +  4  a;"^ 


2afz--2ari 
-2x~i 
2  x~i  -  4  x~Xi  -  3 
-3 

2  af*  -  4  af  *  -  6  -  2  a4 
-2a* 


6aT*  +   8  a;"?  4- 17 
6aT*  + 12  aT4'  +   9 


-  4ari+   8  +  12x£  +  4a^ 

-  4af*  +   8  +  12xT  +  4a^ 


6.   Find  the  value  of  x  in  the  equation  x  *  =  8. 

(Give  to  both  members  the  exponent  necessary  to  make  the  exponent 
of  x  equal  1.) 

aft  =  8,    (art )-f  =  (8)~t,   »  =  8-$,  x  =  — ,   a;  =  J>     Result. 


7.   If  ar3  =  2/-2,  and  2/-1=  —8,  what  is  the  value  of  x  ? 
We  first  require  the  value  of  y~2  from  the  second  equation. 
Hence,  if  y~1  =  -S,  Tnen>  x"3  =  (-8)2. 

(y-i)2zz(-8)2,  *  =  [(-8)2r* 


(-8)2. 


=  -8-1 

=  J.    Result. 


-2m 


234  THEORY   OF  EXPONENTS 

Exercise  80 

Multiply : 

1.  a;"2 +  3  a;"1 -2  by  a;-2 -2  a;-1 -4. 

2.  a*  +  2a±&*  +  6*by  a*-2a*&*  +  &i 

3.  4ar3m  +  6ar2m  —  5arOT-3  by  3 x~2m  +  2 arTO - 1. 

4.  a*-9a*  +  27a~*-27a"Hy  a^-6  +  9a"i 

5.  va^ ^  +  2v^-2V^by  -^  +  Va  +  -^. 

W  Va3  Va 

Divide : 

6.  c-4_c-3_8c-2  +  llc-i_31:?v  c-2  +  2c-i_3. 

7.  2a-a*  +  4a*  +  4a*-3by  a*-a*  +  3. 

8.  35+4a-w-16a-2m+19a-3w-6a-4m  by  7+5orw-3cr 

9.  a2  +  2a;*-7ar*-8af£  +  12ar3by  a?  -  3  oT*  +  2  af*. 

10.  a-M?+^-8K+H7V-a-^+±. 

■y/c        Vc        Vc3      c  Vc      Vc 

Simplify : 

11.  [(x  +  3)(x-3)-1-(x-3)(x+3)-1]  +  [l-(xi+9)(x+3)-2]. 

12.  [(o^-SXc-^-^^^^]  x  [Sc^-O)-}-1 
Expand : 

/      2V       17,    I — = F/ 

14.   (V^+2^)\     16"   (3ViF-aJ'  W°     2V^ 

Extract  the  square  root  of : 

18.  4  ar4  + 12  ar3  +  ar2- 12  a;-1 +  4. 

19.  9a-2"l-6a-w-ll  +  4am  +  4a2w. 


APPLICATIONS  OF  PRINCIPLES  OF  EXPONENTS      235 


20 


.   x~%  —  4  x  ^  +  8  x  %y*  —  8  x~*yl  +  4  y^. 


*      -y/x3        \y  \x 

Find  the  value  of  x  in : 

22.  a£==3.  25.   y~*  =  4.  28.    #"*=£. 

23.  a;*  =  2.  26.    2"*  =  9.  29.    x»   =2. 

24.  a>*  =  25.  '27.    af*  =  — 8.  30.   xm  =  3n. 

Multiply  by  inspection :  Divide  by  inspection : 

31.  (3 a"3 - 2 a3)2.  35.    (a"3 - 9)  by  (a*  +  3). 

32.  (2  or1 +  3)  (3  ar1  +  7).  36.    (a  -  81)  by  "(a*  +  3). 

33.  (3a-1-2a)(5a"1-3a).  37.    (a-3-86"8)by (a-1-26~1). 

34.  (2a-2  +  a~1H-3)2.  38.    (27a"*+125)by(3a~J+5). 

Find  the  value  of  a;  in  each  of  the  following : 

39.  x'1  =  y,  and  y2  =  9.  42.   a£  =  y_1,  and  y  =  3. 

40.  sb  =  2/-1>  and  y*  =  3.  43.    a£  =  2/^,  and  y»  =  9. 

41.  x~3  =  ?/,  and  y~2  =  2.  44.   aT*  =  IT3,  and  y~*  =  9. 

Simplify : 

45.  (a"1  -  or1)2  -  (a"1  +  ar1)  (a"1  -  x-1). 

46.  (a-ffl  +  3am)2-(a-m-3aw)2. 

47.  (a-3  +  l)a-3-(a-3-l)2  +  (l-a-3). 

V      d»-c*A      c^  +  dU    J___L 

49    f     ^  + 13 ajy*  a^  +  4yH  .  |"5^  +  2^"| 

L^4-2a-3^  — 3^      a*  +  3^J     L*t— 9yt.J 


CHAPTER  XX 
RADICALS.    IMAGINARY  NUMBERS.    REVIEW 


252.  A  radical  expression  is  an  indicated  root  of  a  number  j 
or  expression. 

Thus  :  V2,  fyl,  VlO,  and  Vx  +  1  are  radical  expressions. 

253.  Any  expression  in  the  form  -\/#  is  a  radical  expression, 
or  radical.  The  number  indicating  the  required  root  is  the 
index  of  the  radical,  and  the  quantity  under  the  radical  is  the 
radicand. 

In  \/7,  the  index  is  3,  and  the  radicand,  7. 

254.  A  surd  is  an  indicated  root  that  cannot  be  exactly  ob- 
tained. 

255.  A  radical  is  rational  if  its  root  can  be  exactly  obtained, 
irrational  if  its  root  cannot  be  exactly  obtained. 

Thus :  V25  is  a  rational  expression  ;  VlO  is  an  irrational  expression. 

256.  A  mixed  surd  is  an  indicated  product  of  a  rational 
factor  and  a  surd  factor. 

Thus  :  3 V6,  4V7x,  abVa  +  b  are  mixed  surds. 

257.  In  a  mixed  surd  the  rational  factor  is  the  coefficient  of 
the  surd. 

Thus  :  In  4VEx,  4  is  the  coefficient  of  the  surd. 

258.  A  surd  having  no  rational  factor  greater  than  1  is  an 
entire  surd. 

Thus  :  Vbac  is  an  entire  surd. 

236 


TRANSFORMATION   OF   RADICALS  237 

259.  The  order  of  a  surd  is  denoted   by  the   index   of  the 
required  root. 

Thus  : .  a/5  is  a  surd  of  the  second  order,  or  a  quadratic  surd. 
\/7  is  a  surd  of  the  third  order,  or  a  cubic  surd. 

260.  The  principal  root. 


Since  (+  a)2  =  -f-  a2  and  (— a)2  =  -f  a2,  we  have  V+a2=  ±  a. 

That  is,  any  positive  perfect  square  has  two  roots,  one  +  and 
the  other  — ,  but  in  elementary  algebra  only  the  -f  value,  or 
principal  root,  is  considered  in  even  roots. 

THE  TRANSFORMATION  OF  RADICALS 

TO  REDUCE  A  RADICAL  TO  ITS  SIMPLEST  FORM 

261.  A  surd  is  considered  to  be  in  its  simplest  form  when 
the  radicand  is  an  integral  expression  having  no  factor  whose 

;  power  is  the  same  as  the  given  index.     There  are  three  com- 
mon cases  of  reduction  of  surds. 

(a)  WJien  a  given  radicand  is  a  power  whose  exponent  has  a 
factor  in  common  with  the  given  index. 

By  Art.  244,  tfp  =  a*  =  a%  -  Va. 

Hence,  to  reduce  a  radical  to  a  radical  of  simpler  index : 

262.  Divide  the  exponents  of  the  factors  of  the  radicand  by  the 
index  of  the  radical,  and  write  the  result  with  the  radical  sign. 

Illustration : 

V%ahfi  =  y/WahP  =  2yaM  a  Wcfix-    -Result. 

Exercise  81 

Simplify : 

1.  -v/aV.  5.  VW&.  9.  a/243. 

2.  -Vote*.  6.  VWm*.  10.  J/WaW. 

3.  vW.  7.  ^36.  11.  \/ffiS. 

4.  tyrftf*.  8.  -y/lSS.  12.  ^216  aV. 


238         RADICALS.     IMAGINARY   NUMBERS.     REVIEW 

(b)  When  a  given  radicand  has  a  factor  that  is  a  perfect  power 
whose  exponent  is  of  the  same  degree  as  the  index. 

By  Art.  244,  Vcfib  =  (a2&)*  =  aM  =  a$  =  aVb. 

Hence,  to  remove  from  a  radicand  a  factor  of  trie  same  power 
as  the  given  index : 

263.  Separate  the  radicand  into  two  factors,  one  factor  the 
product  of  powers  whose  highest  exponents  are  multiples  of  the 
given  index.  Extract  the  required  root  of  the  first  factor  and  write 
the  result  as  the  coefficient  of  the  indicated  root  of  the  second  factor. 

Illustrations : 

1.    vW  =  V4 a2  x  3 a  =  2 aVSa.    Result. 


2.  2V72  aWy1  -  2  V36  a2x*y«  -2  ay  =  2(6  ax2ys)  y/2  ay 

=  12  ax2y*  y/2ay.     Result. 

Exercise  82 
Simplify : 

1.    V28".  8.    VI§2.  14.   |V*p; 

2.  vs:  9  2*/mu  «•  1^375- 

3-  ^*  io.  3VI62:         16-  *W 

4.    V98T  17.    V9  a3. 

11.   3^80. 


5.    </W.  XX'    ovo^_  18.    Vl6aY. 

e.  VIM  12-  AV720.  i9.  ivTe^: 

7.    a/54.  13.    2-J/243.  20.    </54aV. 


21.  -i-Vl62^y.  23.    |^-V54  mV 
3  y2  9  m2 

22.  V675mV.  24.    ■y/4S<*<P1aP. 


TRANSFORMATION    OF   RADICALS  239 


»■  &   -  m-   ■•  f#§ 


6  3jl25^fz  16&  3/^686" 

'    *  \  216  m3  '  30,      14    \125z6' 


29.    J2B0  («  +  «)«  31.    ((B  +  1)  'I 


i* 


a  +  1)4 

(c)  TT^en  ^e  given  radicand  is  a  fraction  whose  denominator 
is  not  a  perfect  power  of  the  same  degree  as  the  radical. 

If  the  denominator  of  a  fractional  radicand  can  be  made  a 
perfect  power  of  the  same  degree  as  the  index  of  the  radical, 
the  fractional  factor  resulting  may  be  removed  from  the  radi- 
cand as  in  the  previous  case.  By  multiplying  the  denominator 
by  a  particular  factor  we  produce  the  desired  perfect  power. 

Multiplying  both  numerator  and  denominator  by  this  particular 
factor  introduces  1  under  the  radical,  and  the  value  of  the 
radicand  is  unchanged. 

Ingeneral:  ^.^-^^g^ily^ 
Illustrations : 

Hence,  to  reduce  any  fractional  radicand  to  an  integral 
radicand : 


.  Multiply  both  numerator  and  denominator  of  the  radicand 
the  smallest  number  that  will  make  the  denominator  a  perfect 
power  of  the  same  degree  as  the  radical.  By  the  method  of  the 
preceding  case  remove  the  fractional  power  thus  formed. 


240        RADICALS.     IMAGINARY  NUMBERS.     REVIEW 
Exercise  83 

2 


Simplify : 

,4 

>-4 

W! 

10.    M 

,4 

11.  4—- 

*a§ 

12.  J*. 

-4 

13.  mM 

>-4 

14     c    4/^\ 

SC  Ac3 

-4 

K     2c   s/T 
15.    — a  • 

3  V'2c 

•■# 

-  6*VS- 

17. 

m 


3/54  m4 


18.  J-1% 

*#  + 1 

19.    (l_a)^f-— 


20.  bi^SS 

a  +  1  *#  — 1 


21         1       3/(m-iy(m  +  2)6 
m  +  2\         (m  +  1)2 


22       m*       p^-12a;  +  38 
^  —  ft  \ 


a?  —  o  *  fir 


23.    (a-2)xf^± 


a  — 1    w 


J^+2^+1 


TO  CHANGE  A  MIXED  SURD  TO  AN  ENTIRE  SURD 

The  process  is  the  reverse  of  that  of  Article  261,  Section  (a). 

In  general :  ay/x  =  a  •  x%  =  a%2  =  Va2#. 

Hence,  to  change  a  mixed  surd  to.  an  entire  surd : 

265.  liaise  the  coefficient  of  the  surd  to  the  same  power  as  the 
degree  of  the  radical,  and  multiply  the  radicand  by  the  result. 
TJie  indicated  root  of  the  product  is  the  required  entire  surd. 

Illustrations : 

1.  3  V5  =  VPT5  =  \/9^5  =  V45.     Result. 

2.  2  VI  =  \f¥Ti  =  %%7i  =  ^32-     Result. 


TRANSFORMATION  OF   RADICALS  241 


Exercise  84 


Change  the  following  to  entire  surds : 

1.  2V3. 

2.  5V2. 

3.  3^/4.  u    3™    /~8^ 


»X  2 


4.2^2.  2a;V27TO 

5.  3^4.  12.  i^Vg. 

a    *ar 

6.  2xV2x.  , 

7.  fcrfffii  13'    (a  +  1)\(^i)^- 

8.  So2^*;. 


14.    g±Sj      ^-1 

»  — 1  ^arJ  +  4a;H 


9.   3a-\/3a;2.  '    a-1  \ar2  +  4a;  +  4* 

TO  CHAISE  RADICALS  OF  DIFFERENT  INDICES  TO  EQUIVALENT  RADICALS 
HAVING  THE  SAME  INDEX 

We  have  Va  =  a? 

and  -\/a  =  a5. 

Expressing  the  exponents  as  equivalent  fractions  having  a 
common  denominator,  a%=a?  =  -\/a3, 

•  a*  =  a*  =  -v'a*. 

Hence,  to  change  radicals  of  different  indices  to  equivalent 
radicals  having  the  same  index : 

266.  Express  the  given  radicals  with  exponents  in  fractional 
form  and  change  these  fractions  to  equivalent  fractions  having  a 
common  denominator.     Rewrite  the  results  in  radical  form. 

Illustrations : 

1.   Change  V5  and  VlO  to  radicals  having  the  same  index. 

V5    =    5^  =    5*  =  #5*    =  #125. 


#10  =  10^  =  10^  =  #102  =  #100. 

SOM.    EL.    ALG. 16 


Result. 


242        RADICALS.     IMAGINARY  NUMBERS.     REVIEW 

2.   Arrange  ^11,  V5,  and  V90  in  order  of  magnitude. 

vTl  =  113  =  ni  =  ^n-2  =  ^121. 
V5  =   5*  =   5^  =  \/p   =  vT25. 

^90  =  90^  =  #90. 

The  order  of  magnitude  is,  therefore  :  V5,  vTI,  #90.    Result. 

Exercise  85 

Change  to  equivalent  radicals  of  the  same  index : 

1.  V2,  VS.           4.    -y/n,  -v/6.  7.  V2,  #3,  ^5. 

2.  V3,  -Vl.           5.    ^6,  ^/8.  8.  V5   </9,  ^15. 

3.  </2,  -VS.           6.    >^^5.  9.  y/l,</7,  VS. 
Arrange  in  order  of  magnitude : 

10.  V5,  -v^IO.       12.    \/iJ,  VK  14.  V3,  -v/8,  -v^5. 

11.  -v/4,  V3.         13.    -\/5,  </4.  15.  V2,  -#6,  ty^3. 

OPERATIONS   WITH  RADICALS 

267.  Kadicals  that,  when  reduced  to  simplest  form,  differ 
only  in  their  coefficients  are  similar  radicals. 

ADDITION  AND  SUBTRACTION  OF  RADICALS 

268.  Similar  radicals  may  be  added  or  subtracted  by  add- 
ing or  subtracting  their  coefficients. 

Illustration : 

Find  the  sum  of  2Vl2  +  2 V27  -  9 V^~+  5 V3. 

2vT2=       4V3 

o-v/97  —        fiVs  "  tne  &*ven  radicals  are  dissimilar,  such  as 

~~  are  similar  may  be  added  as  illustrated,  theex- 

~~  9v^  *  =  —  12V3        pression  for  the  sum  being  indicated. 
6V3    =        5V3 


Sum        =        3V3    Result. 


OPERATIONS  WITH  RADICALS  243 

Exercise  86 

Simplify : 

1.  VI8  +  V8-V32.  5.    ^2-^/16  +  ^250. 

2.  V75-V48+V27.  6.    ^  _ -</512  +  Vl62. 

3.  2V3-2V27  +  2V108.  7.    V|  +  V6- V|-iV6. 

4.  3V98-2V75-3V32.  8.   V|-  Vf +iVf-iVff. 
9.   V45~S-  V20o¥-  V5~a¥-  V80o*a;  +  V180  a2s. 

10.  V8aa  +  V32aa  —  f  Vl8 ax  —  V16 ax. 

11.  2V7- V80  +  V63-Vil2  +  Vi5. 

12.  4\/24  -2^/81  +  11^3-3^192. 

13.  10Vi-5VJ|  +  24Vi-7V2  +  16VS. 

14.  3^^-4^  +  -\/l28-2^|-  +  6^/J-9^/|. 

15.  8V|  +  14V|-V75-iV5i2  +  5V||-6V||. 

MULTIPLICATION  OF  RADICALS 

269.   Any  two  radicals  may  be  multiplied. 
Illustrations : 

1.  Multiply  3V6  by  2V15. 

(3>/6)(2 Vl5)  =  (3  •  2)( V6  .  Vl5)  =  6V90  =  18\/l0.       Result. 

2.  Multiply  V28  X  V42  X  Vl5. 
Expressing  each  radical  in  prime  factors, 

V28)(V42)(Vl5)  =  V(7.22)(7-2.3)(5.3)  =  V(72  •  22 .  32)  (5  . 2) 
=  (7-2.3)  VTO  =  42  VlO.     Result. 

3.  Multiply  2-v/4  by  3</8. 
Changing  to  radicals  of  the  same  index, 

2^4  =  2  •  4*  =  2  .  4T*  =  2\/4*. 
3\/8  =  3  .  8*  =  3  •  8T2  =  3y/&. 
(2  y/i*)  (3  v/p)  =(2-3)  (  \/5*7p)  ==6v/2^^=6\/2^=6v/2i226".  =12^32. 


244         RADICALS.     IMAGINARY   NUMBERS.     REVIEW 

Hence,  to  multiply  two  or  more  monomial  radicals : 

270.  Multiply  the  product  of  the  given  coefficients  by  the  prod- 
uct of  the  given  radicands,  first  changing  the  radicals  to  radicals 
having  a  common  index. 

The  principle  is  in  no  way  changed  in  application  to  frac- 
tional radicands.  It  is  usually  best  to  multiply  fractions  in  the 
form  given,  reducing  the  product  to  its  simplest  form. 


Exercise  87 

Multiply : 

1.  V3by  V6.  5.   |V6by^Vl5.     9.  fV^-  by  |V28f 

2.  Vl2  by  V8.  6.    -\/4  by  ^2.         10.  V35ac  by  VUax. 

3.  2V7by|V21.  7.    V6  by  ^9.         11.  ^Im^byVBlm^. 

4.  3-\/4  by  -v/12.  8.    V5? by  VJf.     12.  J/2nm  by  Vl2msc. 

13.  (2V3  +  3V6-2Vl2)(2V3). 

14.  (4^/2 -3  ^4 +  2  a/12)  (^12). 

15.  (V|  +  iVf-3V|)(V|). 

16.  (2 V6)  (2 V3  -  3 V6  +  i  V15  +  i  VlO). 

MULTIPLICATION  OF  COMPOUND  RADICALS 

271.    Illustration: 

Multiply  3  V2m-2 V6m  by  2V2m  +  3  V6m. 

3  V2ro  -  2  V6m 
2  V2m  +  3  V6m 


6  •  2  m  -  4  Vl2  m2 

+  9  V12  m2  -  6  •  6  m 


12  m  +  5  V12  to2  -  36  m  =  5  Vl2  ra2  -  24  m  -  10  m  V3  -  24  m.     Result. 


OPERATIONS  WITH  RADICALS  245 

Exercise  88 

Simplify: 

1.  (2+V2)(3  +  V2).        5.  (4V3  +  2)(2V3-2). 

2.  (3+V5)(2  +  V5).        6.  (3^3-2-v/9)(2-v'3  +  3-v/9). 

3.  (4  +  2V3)(3  +  2V3).   7.  (4 V5 - 2 V3) (5 V5  +  3 V3). 

4.  (5-2V2)(2-V2).      8.  (2a/4  +  V2)(3^'4-V2). 
9.  (Va+ V^)(Va  + V»). 

10.  (2Vm-3Vw)(3Vm-2Vn). 

11.  (4ajV2-?/V3)(2a;V2  +  2/V3). 

12.  (3V2^-2V5^)(2V2^  +  4V5y). 

13.  (3V3-2V2-4V5)(2V3  +  V2-2V5). 

14.  (3V6  +  V2-V3)(4V2+V6  +  V3). 

15.  (V2  +  V3)(V3  +  V6)(V3-V2). 

16.  (V6-V3)(V2-V6)(V3-V2). 

17.  (V^TT-Va^l)(V^+l-2V^^l). 

18.  (2VaT^-3Va^)(3VaT^  +  2V^T). 

19.  (3Va^  +  a  +  l-2Va;  +  l)(2V^  +  a;  +  l  +  2Va>  +  l). 

DIVISION  OF  RADICALS 

272.   As  in  multiplication  of  radicals  we  may  divide  any 
two  radical  expressions  of  the  same  index. 
Illustrations : 

1.  Divide  -^96  by  y/2. 

M  =  ^.=  m  =  2Vz.    Result. 

2.  Divide  y/l  by  V6. 

n=w=yii=^i^  Result 

V6      W       >216       *27      3 


246        RADICALS.     IMAGINARY  NUMBERS.     REVIEW 
3.   Divide  yfiff  by  >^f. 

Exercise  89 
Divide : 

1.  V8by  Vl2.  5.  6V5by2Vl5.  .    9.  ^36  by  -y/U. 

2.  V6by  V18.  6.  4-^4  by  3^12.     10.  </\  by  VJ. 

3.  V27by  V12.  7.  JV72  by  fVJ.     11.  Vffby^f. 

4.  4Vl5by2V5.  8.  -?^  by  ^50.       12.  V^  by  ^£. 

DIVISION  BY  RATIONALIZATION 

273.  If  a  given  divisor  involves  quadratic  surds  only,  a 
division  is  really  accomplished  by  the  process  of  rationalization ; 
or,  by  multiplying  both  dividend  and  divisor  by  the  expression 
that  will  free  the  divisor  from  surds. 

(a)    When  the  Divisor  is  a  Monomial. 

Illustrations : 

1.  Divide  8  by  2  V3. 

_^  =  _8_XV3  =  8V3  =  8^  =  4V3>     Result>        . 
2V3     2V3      V3     2.3        6        3 

2.  Divide  2  -tys  by  -y/L 

2jft  =  2V8xtt  =  2(lVy^)=2WW)  =  2i^     Regult 

The  process  of  rationalization  of  the  divisor  simplifies 
numerical  calculations  with  radical  expressions. 

3.  Find,  to  the  three  decimal  places,  the  value  of     "Y,  ' 

Vl2 

3V6     3V6xV3     3V18     9V2     3    _     S„„A       N      mmm£ 

—=z  =  —= 7=  =  — =■  =  -7—  =  -0v5  =  o  (1.414+-..)  =  2.121+.... 

Vl2      Vl2  x  V3       V36         6        2  2  ^  > 


OPERATIONS    WITH  RADICALS  247 


Exercise  90 


Eationalize  the    denomi-  Find,  to  three  decimal  places, 

nators  of :  the  value  of : 

1.    — -•      .  4.     -•  7. 10. 


V2      .  SV2  V5  V75 

&  5.  A.  8.  ju  ii.  b* 

V6  V2  3V2  3V2 

-*-.  6.    _L.  9.    ?V5.  12     A. 

2V3  a/9  3V3  ^i 


(5)    When  Either  or  Both  of  the  Terms  of  a  Binomial  Divisor 
are  Quadratic  Surds. 

By  Art.  104,  («  +  6)  (a  -  b)  =  a2  -  6*. 

Similarly,  ( Va  +  V&)(  Va  -  Vb)  =  a  -  b. 

274.  Two  binomial  quadratic   surds  differing  only  in  sign 
are  called  conjugate  surds. 

( Va  +  Vb)  and  ( Va  —  Vb)  are  conjugate  surds. 
From  the  multiplication  above  we  may  conclude : 

275.  The  product  of  two  conjugate  surds  is  rational. 
Illustrations : 

2V3-3V2 


1.   Rationalize  the  denominator  of 


2  V3  -  V2 


2V3-3v^_2V3  -3\/2x2\/3+  V2 
2V3-V2       2V3-V2      2V3  +  V2 

_6-4V6_,2(3-2V6)^3-.2V6      Result 
10  10  5 

Two  successive  multiplications  will  rationalize  a  trinomial 
denominator  in  which  two  quadratic  surds  are  involved. 


248        RADICALS.     IMAGINARY  NUMBERS.     REVIEW 

2.   Eationalize  the  denominator  of — ^t—  • 

V3+  V2-1 

V3_  V2+  1  =  V3-(V2-1)      V3-(V2-1) 

V3+V2-1      V3+(V2-1)      V3-(V2-1) 

=  6-2V6  +  2a/3-2V2* 

2V2 

3  -  V6  4-  V3  -  V2 

V2 

3_V6+V3-V2x  V2_3\^-2V3+  V6  -  2 

V2  V2  2 

Exercise  91 

Eationalize  the  denominators  of: 


4     V2+  V3  7>    2V2-4 


2  +  V2  V2-V3                     3V2+2 

3  5     5-V2  g     4V3-3V2 
V2-2*  '    2  +  V2*  '     2V3  +  3V2' 
5_  3V2  +  1  9     5V6-2V3 


V7-V2  V2-1  3V2  +  3 

10     V^+V^-  15.     Vs  — 2  —  Va^ 

Va—  V#  V#  —  2  +  Va; 

„      2Vm+Vn  .        -„     Va  + 1  +  V2  a  —  1 

11«    = r*  ■*■*>.     — rzz= z' 

3  Vm  —  Vw  Va  + 1  —  V2  a  —  1 

2Va  +  V2a,  17.    V3  +  V2-1, 

3Va-V2^  V3-V2  +  1 

13     qVft  +  cVa?  3- V2  + V3 

aVaj  — cVa  3  +  V2  — V3 

V^fl+2  4  -  V  3  +  V5 

14.     —  ■ ly-     — -=• 

Va  +  1-2  4_VB-V5 


12. 


INVOLUTION   AND   EVOLUTION   OF   RADICALS        249 
INVOLUTION  AND  EVOLUTION  OF  RADICALS 

276.   By  the  aid  of  the  principles  governing  exponents,  we 
may  obtain  any  power  or  any  root  of  a  radical  expression. 

Illustrations : 

1.  Find  the  value  of  (2  Va5)2. 

(2^)2=  (2  afy  =  W  =  4  a?  =  4  Vol    Result. 

2.  Find  the  cube  root  of  VaV. 

VVaW  sk  VaW  =  a?x*  =  Vatf  =  xVax.    Result. 


Exercise  92 

3/ — — 


\ 


Find  the  value  of : 

1.    (Va)6.     •  5.    y/^/S.  9.    V^a  +  fc)2*. 

2.  (-^y)3.  6.  </^  io.  i/yfajfy. 

3.    (-^)«  7.    Vj^L  11.    ^vS 

4.  (-V1T0)6.         8.  V^f^.  12.  e^i^ 

PROPERTIES  OF  QUADRATIC  SURDS 


277.   A  quadratic  surd  cannot  equal  the  sum  of  a  rational  ex- 
pression and  a  quadratic  surd. 

That  is,  Va  cannot  equal  b  +  Vc. 

If  b  is  a  rational  quantity  and  Va  and  Vc  are  quadratic  surds, 

If  Va  =  &  +  Vc, 

we  may  square,  and  a  =  b2  +  2  6  Vc  +  c. 

From  which,  2  &  Vc  =  a  —  &2  —  c. 


w 


Or,  Vc      «-&2-c 


c 


26 
That  is,  we  have  a  quadratic  surd  equal  to  a  rational  expression,  an 
impossible  condition  because  of  the  definition  of  a  surd.     Therefore, 

Va  cannot  equal  b  +  Vc. 


250         RADICALS.     IMAGINARY  NUMBERS.     REVIEW 

278.    Given,    a  -f-  V&  =  c  +  Vd,    V&  and   -Vd   6emgr    surds, 
2%eft  a  =  c  and  b  =  d. 

If  a  is  not  equal  to  c,  we  have,  by  transposition, 
a  =  c+  Vd  —  V&, 
which,  by  Art.  277,  is  impossible.     Therefore  a  must  equal  c.     Hence  it 
follows  that  Vb  =  Vd. 


279.    Given,  Va  +  V&  =  Vx  +  Vy. 

7%ew,  V«_  V6  =  V5  —  Vy, 

to^en  a,  6,  a;,  anc?  y  are  rational  expressions. 


Va  +  V&  =  Va-  +  Vy . 
Squaring,  a  +  V&  =  x  +  2  Vsey  +  y. 

By  Art.  278,  a  =  x  +  y.  (1) 

And  Vb  =  2  Vsy.  (2) 

Subtracting,  a  —  V&  =  x  —  2  Vxy  +y. 

Extracting  sq.  rt.,        V a  —  Vb  =  Vx  —  Vy. 


THE  SQUARE  ROOT  OF  A  BINOMIAL  SURD 

The  square  root  of  an  expression  consisting  of  a  rational 
number  and  a  quadratic  surd  is  obtained  by  application  of 
the  principles  of  Arts.  277,  278,  and  279. 


(a)  The  General  Method 
280.    Illustration: 
Find  the  square  root  of  11  +  V96. 


We  may  assume  that,  Vll  +  V96  =  Vx  +  Vy.  (1) 

Then  (Art.  279),  Vll  -  V96  =  Vx  -  Vy.  (2) 

Multiplying  (1)  by  (2),  Vl21  -  96  =  x  -  y. 

Or,  x  —  y  =  5.  (3) 


SQUARE  ROOT  OE  A   BINOMIAL  SURD  251 


(4) 


Squaring  (1), 

11  +  V96  =  x  +  2  y/xy  +  y. 

Hence  (Art.  278), 

11  =  x  +  y. 

From  (3)  and  (4), 

x-y  =  b. 

3  +  ^  =  11. 

Whence, 

x  =  8,  y  =  3. 

Therefore, 

V»  +  y/y  =  V8  +  V3  =  2  V2  +  V3. 

That  is, 

Vll  +  V96  =  2V2+  V3.    Resu 

(b)  The  Method  of  Inspection 

281.   If  a  binomial  quadratic  surd  can  be  written  in  the  form 
a  +  2  V&,  its  square  root  may  be  obtained  by  inspection. 

Illustration : 

1.   Find  the  square  root  of  23  +  6  VlO. 

6  VlO  may  be  written  thus:  6  VlO  =  2  (3  VlO)  =  2  V90. 

Then  23  +  6  VlO  =  23  +  2  V90.     (The  required  form  of  a  +  2  Vb.) 

The  factors  of  90  whose  sum  is  23  are  18  and  5. 

Hence,  the  square  root  of  (23  +  2  V90),  or  the  square  root  of 

(18  +  2  V90  +  6)  =  Vl8  +  V5  =  3  V2  +  V6.     Result. 

Exercise  93 

Find  the  square  root  of : 

1.  7  +  4  V3.  9.  146-56V6. 

2.  11-6V2.  10.  124-30VTL. 

3.  17  +  12V2.  11.  3x  +  2x^2. 

4.  28  +  6V3.  12.  c2  +  m  +  2cVm. 


5.   33-8V2.  13.   2m  +  2Vm2-l. 


6.   30-12V6.  14.   2m2-l  +  2mVm2-l, 


7.  67  +  16V3.  15.   c2  +  c  +  l-2cVc  +  l. 

8.  138  +  30V21.  16.  a2+4a+l  +  (2a+2)V2o: 


252        RADICALS.     IMAGINARY  NUMBERS.     REVIEW 
EQUATIONS  INVOLVING  IRRATIONAL  EXPRESSIONS 

282.  Equations  in  which  the  unknown  number  is  involved 
in  radical  expressions  are  called  irrational  equations. 

283.  An  equation  containing  radicals  is  first  rationalized  by 
involution.  If  there  are  two  or  more  radicals  in  the  given  equa- 
tion, it  may  be  necessary  to  repeat  the  process  of  squaring 
before  the  equation  is  free  from  radicals. 

284.  Roots  of  Irrational  Equations.     It  can  be  shown  that 

(1)  The  process  of  squaring  may  introduce  a  root  that  is  not 
a  root  of  the  given  equation. 

(2)  An  irrational  equation  may  have  no  root  whatever. 

Hence,  we  conclude : 

285.  Any  solution  of  an  irrational  equation  must  be  tested,  and 
a  root  that  does  not  satisfy  the  original  equation  must  be  rejected. 

Illustrations : 


1.   Solve  a>  +  3  =  Vaj"  + 15. 

X  +  3  =  Vx2  + 15. 
Squaring,  x2  +  6  x  +  9  =  x2  +  15. 

Transposing,  x2  —  x2  +  6  x  =  —  9  +  15. 

Uniting,  6  x  =  6. 

Whence,      ,  x  =  l. 

Substituting  1  in  the  original  equation,  and  taking  -the  positive  value  of 
the  square  root,  we  have, 

1  +  3  =  vT+To. 
4=4. 
Therefore,  the  solution,  x  =  1,  is  a  correct  solution. 


2.   Solve  VaJ  —  2  +  Vx  +  5  =  7. 

y/x-2  +  Vx+5  ss  7. 
Transposing,  Vx  —  2  =  7  -  Vx  +  5. 


Squaring,  x-2  =49-14V»  +  5  +  X  +  5. 

Whence,  14  Vx  +  5  =  56. 
Dividing  by  14,  Vx  +  5  =  4. 

Squaring,  x  +  5  =  16. 


EQUATIONS   INVOLVING   IRRATIONAL  EXPRESSIONS    253 

And,  a;  =  11. 

In  the  original  equation,     Vll  —  2  +  Vl  1  +  5  =  7. 

V9+ Vl6  =  7. 
7  =  7. 
Hence,  x  =  11,  is  a  solution  of  the  given  equation. 

3.   Solve  VaT+3  —  5  =  Vx  —  2. 


Vx  +  3  _  5  =  Vx  -  2. 
Squaring,  x  +  3  -  10  Vx  +  3  +  25  =  x  -  2. 

Transposing,  x  —  x  —  10  Vx  +  3  =  -  3  —  25  —  2. 

Collecting,  -  10  Vx  +  3  =  -  30. 

Dividing  by  -  10,  Vx  +  3  =  3 . 

Squaring,  x  +  3  =  9. 

Whence,  x  =  6. 

Substituting  in  the  original  equation, 


V6  +  3  -  5  =  V6  -  2. 

V9  -  5  =  VI. 

3-5=2. 

Since  the  original  equation  is  not  satisfied  by  the  solution  x  =  6,  this 

solution  must  be  rejected. 

It  will  be  found  by  trial  that  the  solution,  x  =6,  satisfies  the  equation 
5  -  Vx  +  3  =  Vx^2 . 

Exercise  94 

Solve  and  test  the  solutions  of : 


1.    Va;+1  =  2.  8.  V7+a;-7  +  VaJ  =  0. 

2.  v^i=3.  9-  Vt+s-vs+4 

3.  V^T  =  ,-1.  10-  2(V^5)(V^-5)=-35. 

11.  Vx  —  1— Va  =  2. 

4.  2VS-3-V5+2.  12  V-_[=V5_TI_V^ 

5.  ^a?-l  =  V6.  13  3V^+V^=2=  V4»-l. 

6.  2VaJ+3=VaJ  +  2.  14.  V#  +  a  +  Va  4-  2  a  =  V4  a;  —  a. 


7.    V«2  +  3  =  a;-l.         15.    yz-5-V4a;-2+-  Va  +  3  =  0. 
16.    V4  a  +  c  =  V25  x  —  3  c  —  V9  <c  —  c. 


254         RADICALS.     IMAGINARY   NUMBERS.     REVIEW 


17.    V36a-1  =  V4a-1  +  Vl6a; 


18.   Va  + 1  +  V5  +  2  =  V5  —  1  —  Va;  —  3. 


19.    VaJ-5  + Va  +  8  =  Va?-3  +  Vaj  +  2. 

20.    — —-VC^-Vi  22.    V6  +  V5  +  V^  =  3. 

Va?  — 7 

2i.  vss+i — -Jg-^vssL    23.  Vfl;-9+v^=-i. 

V2a+1  VaJ-9— Va: 

24.  J*±|_J*E|  =  0. 

2V^+3     4VaT+l     A 

25.    7= 7= =  v. 

3Va;-l      6Va;-2 

26.  V?a-V3»=V3a  +  2Vo[6aj-(4-a)]. 


IMAGINARY  AND   COMPLEX   NUMBERS 

286.  The  factors  of  -fa2  are  either  (+a)  and  (+a),  or 
(—a)  and  (—a).  The  factors  of  —a2  are  (4- a)  and  (—a). 
Clearly,  therefore,  no  even  root  of  a  negative  number  is  pos- 
sible.    Hence : 

287.  An  imaginary  number  is  an  indicated  even  root  of  a 
negative  number. 


V—  2,  V—  5,   V—  10,  are  imaginary  numbers. 

288.  The  symbol,  V^l,  is  the  unit  of  imaginaries. 

289.  The  rational  and  irrational  numbers  hitherto  con- 
sidered are  known  as  real  numbers  in  contradistinction  to  this 
new  idea  of  imaginary  numbers. 

290.  Imaginary  numbers  occurring  in  the  form  of  V—  b, 
where  b  is  a  real  number,  are  pure  imaginaries. 


OPERATIONS  WITH  IMAGINARY  NUMBERS  255 

291.  An  imaginary  number  occurring  in  the  form  of 
a-f-V—  b,  where  a  and  b  are  real  numbers,  is  known  as  a 
complex  number. 


The  Meaning  of  a  Pure  Imaginary. 

By  definition,  Va  is  one  of  the  two  equal  factors  of  a.     Or, 

(v^)2  =  VJaj*      =  («)t      =  a. 
In  like  manner,  (  V^l)2  =  V(-  l)2  =  (-  1)1  =  -1. 

OPERATIONS   WITH   IMAGINARY  NUMBERS 

293.  The  fundamental  operations  with  imaginary  numbers 
are  based  upon  the  principles  governing  the  same  operations 
with  radical  expressions. 

294.  A  pure  imaginary  number  in  which  the  real  factor  is  a 
perfect  square  may  be  changed  to  the  form  aV—  1,  in  which 
form  the  elementary  processes  involving  imaginary  numbers 
are  greatly  simplified. 

Thus,         (1) *       V^"4  =  V(4)(-l)  =  2  V^l. 

(2)  2\^l9  =  2V(49)(-  1)=  14\/^I. 

(3)  av^x*  =  aV(x*)(-l)  =  ax2 V^T,  etc. 

ADDITION  AND  SUBTRACTION  OF  IMAGINARY  NUMBERS 

We  assume  that  the  principles  underlying  the  addition  and 
subtraction  of  real  numbers  will  still  hold  true  in  the  addition 
and  subtraction  of  imaginary  numbers. 

Illustration : 

Find  the  sum  of  5V^9  +  \f^2&  -  3  V^IB. 
5V^9  =    15V^T 
v^26=       5v^T 
-3V^l6=-12\/^T 

Hence,  8  V—  1     Result. 


256 


RADICALS.     IMAGINARY  NUMBERS.     REVIEW 


Exercise  95 


Find  the  sum  of : 
i.  V^  +  v 

2 


9  +  V-16. 


V^i9  _  V- 25  +  V- 36. 


3.    V-81  +  V-64 


100. 


4    2V-25-3V-49-2V-64  +  3V-81. 
5.   4V:rI-5V^16  +  3V^25  4-iV:r64. 


6.    V^^  +  V^Ta2-  V- 16 a2- V- 25a2. 


7.    v^l^-  V-(a  +  2)2+V-4. 
2V^" 


8.   3V^i  +  5V-2V--  v-T 

-L         /  9    9  -L         /  9   79  -1- 


+  6V- 


1 
ST' 


1 


9.   -y/ZTM-±^-c>d?-±^-a2<?  +  ±V-aW. 
a  d  c  d 

2    > 3 


10.  cV-l-f- V-16c2+ V-c2 

c 

11.  (2  +  5V^)  +  (7-2V^)-(8-2V^T). 

12.  (x  +  y^/^)  +  (x-z^~^)  +  (y-x->J^l). 

MULTIPLICATION  OF  IMAGINARY  NUMBERS 

295.    The  positive  integral  powers  of  the  imaginary  unit, 

By  Art.  292,  (V^T)2  =- 1. 

Therefore,       (V^H)3  =  (\/^T)2(\/^T)  =(-  1)  (V^I)  =  -  \T^\. 

(V31)4=(V31)2(V31)2=(_1)(-1)  =  +1^ 

(V^T)^(vr^l)HV^l)  =  (+i)(V^I)-vr-i. 

That  is,  the  first  power  of       V—  1  =  V—  1. 

the  second  power  of  V—  1  =  —  1. 

the  third  power  of     V—  1  =  —  V—  1. 

the  fourth  power  of  V—  1  =  1. 

And  this  succession  repeats  itself  in  order  for  the  following  higher 
powers. 


OPERATIONS   WITH   IMAGINARY   NUMBERS 


257 


In  the  multiplication  of  imaginary  numbers  it  is  helpful  to 
remember  that 

296.  The  product  of  two  minus  signs  under  a  radical  is  a 
minus  sign  outside  the  radical. 

Illustration : 

1.   Multiply  V^3  by  V^6. 

VZ3 by  V^6  =  V(-3)(-6)=  V(18)(-l)2=  -  Vl8=  -3  V2.    Result. 


Exercise  96 


Multiply : 

1.  V^  by  V" 

2.  V^G  by  V- 


3.  V-  12  by  V-  3. 

4.  V^36by  -V^16. 


5.  _V-18by  -V-54. 

6.  -yf^-2  by  V^^T  by  V- 

7.  -yf^l  by  V^9  by  V- 

=  »)( 


28. 


16. 


(• 


V- 25)  (V-ioo). 


9.    (V3^)(_V-4c2)(- V-16  62). 

10.  (-V^^C-aV^H-aV^^. 

11.  (-V^l)  (2V^1)  (-3V=i). 

12.  (-2V"=l8)(-3V":r8)(5V^:32). 

13.  (V^aV)  (- -V^^x)  (  -  V  -  c2x). 

14.  (a V"11^)  (a"1-/-!)  (  -  a-V^l). 


15.  (—  Vm  —  »)  ( vn  -  to). 

16.  (V-^-2x-l)  (V^I)  (VT=^). 

MULTIPLICATION  OF  COMPLEX  NUMBERS 

297.  We  assume  that  the  principles  underlying  the  multipli- 
cation of  real  numbers  still  hold  true  in  the  multiplication  of 
complex  numbers. 

SOM.    EL.    ALG. 17 


258 


RADICALS.     IMAGINARY   NUMBERS.     REVIEW 


Illustrations : 

1.   Multiply  3  +  V^2  by  2  -  V^2. 

2  -  Bv^-2 


6  +  4V^2 
-  15V^2-10(~2) 
6-llV^2  +  20  =  26-  11V^2.     Result. 

2.  Expand  (3  -  V^)4. 

In  processes  involving  frequent  repetitions  of  the  imaginary 
unit,  V—  1,  it  is  convenient  to  substitute  t  in  place  of  V—  1, 
replacing  the  imaginary  unit  after  the  simplification  is  com 
plete. 

Let  (3  -  v/^T)*  =  (3  -  iy.     Then  : 
(3  -  iy  =  81  -  108  i  +  54  &  -  12  0  +  & 

=  81  -  108V^T  +  54(-l)  -  12C-V~T)  +  (1)     (Art.  292) 

=  81  -  108v^I  -  54  +  12V^1  +  l 

=  28  -  96  V^l.     Result. 


Simplify 


Exercise  97 


3). 


1.  (2  +  V- 1)  (3  +  V- 1). 

2.  (3-V^l)(4-V^l). 

3.  (2  +  2 V^T)  (3  +  2V"=:T). 

4.  (V^+V^)(V^T+V- 

5.  (2V:::l-3V:r3)(2V":rT- 

6.  (3 V2  +  2 V^5)  (2 V2  -  3V^ 

7.  (V—  a  —  x)  (V—  a  +  a). 

8.  (-V^l  +  2 V^2  +  3 V"^)2. 

9.    (2  +  3V^I)8.  11.    (3V2  -  2V=T)8. 

10.   (3  +  2V-T)«  12.    (aV^T-  bV^iy 


-2). 

4V" 

rg). 


OPERATIONS   WITH   IMAGINARY  NUMBERS 


259 


DIVISION    OF  IMAGINARY  AND   COMPLEX   NUMBERS  BY  RATIONALIZATION 
OF  THE  DIVISOR 

298.   Illustrations: 

1.  Divide  V- 28  by  V^8. 

■ =  )    . {}   -==±  =  \T  =  \T  =  5 VIl.     Result. 

2.  Divide  V- 12  by  V3. 

yrrT2  =  (V—2)xVs=V^==6V=T=2Vzr-v    Regult 
V3  V3xV3  3  3 

3.  Divide  2  +  2V:r3  by  2  -2V:r3. 
2  +  2V^3     2  +  2 V^l?  =  (2  +  2\Z^3)«  _  4  +  8 V^3  -  12 


2  +  2V-3     22-(2V-3)2 
=  8V~3-8_- 
16 


4  + 
"3-1 


Result. 


Exercise  98 


Simplify 
5 


2. 


V^2 

V18 

V^3* 


V-12 


4. 


13 


V-12 
2V^3* 
2V^3- 


5. 


3V- 

x  + 


3+V- 


«-V-l 


V-27 

2V^~3 

2V^~4 

-SV^ 

-14 

V^2 

-4V20 

V-5 
15. 

16. 


10. 


11. 


12. 


5-V-2 


2+V- 
3+V" 


3-V-3 

V2+V^T 

V2-V^l* 

V^2~+V^3 


V-2-V-3 
Vc  +  3V-2c 

Vc—  V—  c 
2  m  -  3  #  V^^T 


2m 


-1 


260 


RADICALS.     IMAGINARY   NUMBERS.     REVIEW 


299.  Conjugate  Imaginaries.  Two  complex  numbers  differing 
only  in  the  sign  of  the  term  containing  the  imaginary  are  con- 
jugate imaginary  numbers. 

Thus,  a  +bV—  1  and  a  —  by/—  1  are  conjugate  imaginaries. 

By  addition,  (a  +  b  v^T)  +  (a  -  6  V^l)  =  2  a. 

By  multiplication,         (a  +  b 


l)(a 


1)  =  a2  +  &2. 


300.    T7ie  sum  and  the  product  of  two  conjugate  complex  num- 
bers are  real. 


MISCELLANEOUS  PROCESSES  WITH  IMAGINARY  NUMBERS 

Exercise  99 
Simplify : 

i.  (V"^i)5  +  (V=T)3.  4.  (_V^I)5  +  (-V^i)4. 

2.  (_v^T)3-(-V"=T)2.      5.  (_V^2)2-(-V^2)4. 

3.  (_2V^2)2  +  3(V^3)3.       6.    (V^I-1)2-(1  +  V^T)S 

7.  (V"=3-l)4. 

8.  (1  +  V"^T)3  +  (l  -  V^T)2  -  (1  -  V^l). 

9.  3(V^)3-3(V"^)(V^-1)2. 
10.   (m  +  w V—  l)(m  —  nV- 1). 


11. 


12. 


13. 


V- 

-1+1 

1 v: 

-i-i 

=1-1 

1        _ 

ri+i 

(V 

:r2)3  + 

(V- 

^y. 

14. 


(2  +  V-l)2  +  (2- 


1£ 


3V-1 


15    a  +  #V—  1      a  —  x 


1 


2V-2  +  V-1 
(V3i)4,(V=2)4 

yzri_V-2 


a  —  xV—l      a  +  x-V—  1 

i6.  ft^i-ff-v=i; 


17. 


V-i 

( V3l)6  _  ( V3T)5  +  ( V3l)4 

V^2 


GENERAL  REVIEW  261 

GENERAL    REVIEW 
Exercise  100 

1.  Expand  (xk  -  2  -s/xf. 

2.  Find  the  square  root  of  (x2  +  3  x  +  2)(x*  —  l)(a?  +  x  —  2). 

3.  Simplify  Vf  +  V^f  +  V|  -  VJ  -  V8. 

5.  Factor  27" V  -  8"1. 

6.  Find  the  approximate  numerical  value  of 

2V2  +  1  .  3V2-1 
V2-1   I    V2  +  1  ' 

[    Simplify  (^-y(^)« 1 

8.  Soivef^-^  =  w2-^ 

[  my  —  %#  =  raJ  —  w . 

9.  Find  the  square  root  of  a*  +  a*  —  4  a6  -f-  4  a  +  2  a«  — 4  al 
10. 


|  simpiify(V^W^)2 


11.  Simplify  (-  V^)3  -  (-  V^T)4  -  (-  V-  l)5. 

12.  How  may  the  square  root  of  (x2— x— 2)(x2— 4)(arJ+3a;+2) 
be  found  without  multiplying  ? 

i(*  +  2/)-z  =  -3, 

13.  Solve  J  2x  +  (z-y)  =  9, 

_z  +  \(x  +  y  +  z)  =  2. 

14.  Factor  a;  (x  +  1)+  a  (2  a?  +  1)  +  a2. 

15-  Simplify  ,  ^C^%#£±gg    • 

1     J  (x-l)-1-(x-2)-1-(x-3)-1 


262         RADICALS.     IMAGINARY  NUMBERS.     REVIEW 


1 


16.  Simplify  <?*+*>  •  — 

17.  Simplify(V^-V^-V^3)(V^I+V^2-V^3). 

18.  Solve  x  —  2y  =  3f  3y  +  z  =  ll,  2z  —  3x=l. 

19.  Simplify  [^.^B4. 

L    ar1  V  a-1        a"2  V  ar3J 

20.  Draw  the  graphs  of  2  a?  +  3  y  =  13  and  3  a?  —  y  =  3,  and 
check  by  solving  the  system. 

21.  Solve  5x  +  2y  —  z  =  3x  +  3y  +  z  =  7x  +  4:y  —  z  =  3. 

22.  Factor  ar4  —  4ar2  —  21. 

23.  What  is  the  square  root  of  x  (x3  -  4)  + 1  -  2  x2  (2  x  -  3)  ? 

24.  Find  the  cube  of  -  2  V^l  +  (V^I)"1. 

rixl6«x2*l-i; 

25.  Simplify  Li___J_^. 

26.  Simplify  cc  +  [x  +  a?  (as  +  ar1)"1]-1. 

27.  Simplify  (-  2  V^  -  V"^)  (3  V^J.+  V-2). 

28.  Expand  (2  a2 -3  a;)5. 

«~     o  i      J?  j  ra  — n  +  1         2  „    ,  3m+n- 1     r 

29.  Solve  for  m  and  n : ! — -  =  —  and —  =$ 

2m  +  n—l      13  m  — ti  +  2 

30.  Simplify  g^+g^i; 

31.  Factor  c(<*  -f  »  +  2 c)  +  c(a?  + 1)  +  (»  + 1)* 

««ai      3      4      1    4  .  2      06,8      1 

32.  Solve =  1,  -  +  -=3,  -  -f  -  =  1. 

x     y  x     z  z      y 

33.  Explain  the  meaning  of  a~n  —  —  • 


GENERAL   REVIEW  263 

34.  Simplify  —  -  3  x°  +  27"^  - 13*. 
8"~* 

35.  Simplify  16c4  +  8c3-2c-M"2-l.fl L^\ 

J  l-9a2  4c2  +  2c   ^     1_2-^V# 

36.  Solve  for  s :  V  4  s  —  9  —  Vs  +  1=  Vs  —  4. 

37.  Find  two  numbers  whose  sum  is  m  and  whose  differ- 
ence is  n. 


38. 


Simplify  |^+flz£ .  (c-Q(l  +  c-i)T 


39.  Find  to  three  decimal  places  the  value  0f2^+3V2t 

3V3-2V2 

40.  Solve  2x  =  5y,  y-X-2z  =  l,  12y-3  +  z  =  4:X. 

41.  Simplify  - ^-l-f. 


2+V-l      5     2-V-l 
42.   Find  the  square  root  of  1~^(2~^) 1£ 


43.  Is  the  expression  a4x  +  2  a3*  +  a2x  +  2  a*  -{-  2  +  a"2*  a  per 
feet  square  ? 

44.  Simplify  2m(m~1)  + L_  . 

(l_m2)f         (l-m2)i  ^ 


45.   Factor 1 +  2(3  +  *>  +  ?*±i. 

(a2^^)-1^     a-1      ^     2-1 


af) 

46.  Solve  -i  +  -i  =  m,—-i  =  n.       • 

ex     ay  dx      cy 

47.  If  a~%  =  c"1,  and  c*  =  f,  find  the  numerical  value  of  a. 

48.  Simplify 

Y— L^-f      ^Y— -i L^ 

V3+V-1   3-V-1A3+V-1   3-v"^iy 

49.  Collect  3Vl|-4V3-6V5|  +  9V8j. 


264         RADICALS.     IMAGINARY   NUMBERS.     REVIEW 

50.  Under  what  condition  is  a  solution  of  mx  +  ny  =  a  and 
kx  +  ly  =  b  impossible  ? 

51.  What  must  be  the  values  of  m  and  n  in  order  that  the 
division  of  x4  +  x8  +  mx2  +  nx  —  3bj  x2  —  x  +  1  may  be  exact  ? 

52.  Extract  the  square  root  of 

4  a2n  -  24  a~n  +  9  a~2n  - 16  an  +  28. 


53.  Simplify  V(a  +  l)(a>*- 1)  -  Va^-ic2-  VaT^l. 

54.  A  discount  of  10  %  was  given  on  a  bill  of  rubbers,  and 
one  of  5  %  on  a  bill  of  shoes.  The  amount  of  both  bills  was 
$  250,  but,  with  discounts  off,  the  merchant  paid  both  with  a 
check  for  $230.     What  was  the  amount  of  each  bill  ? 

55.  What  is  the  value  of  2  a°[2°+(2  a)°]  ? 

56.  Simplify 

[ab-h-1  +  a-'bc-1  +  a^b^c)  '  \   (a  +  b  +  c)~2  J ' 

57.  Expand  |>  +  xl  —  2]3. 

58.  Draw  the  graphs  of  x-\-3y  —  12  and  3x  —  4?/ =  10. 
Checlf  by  a  solution. 

59.  Calculate  to  three  decimal  places  the  value  of 

(1  -  V3)  +  (1  +  V3). 

60.  Simplify    °-°~1       a  +  a_1 


a*  —  a  »     a*  +  a"J 

61.  Solve  («  +  l)(»-l)->-  — — 1— -—  =  8ar>. 

62.  Introduce   the   binomial    coefficient   under   the   radical 
and  simplify  :  (V3  -  2)V7  +  4V3. 


' 


GENERAL  REVIEW  265 

63.   Simplify  1«-  -  3  aP+ 3  (3  x)°  - 1°  +  (1  +  a°)(l  -  a0). 

64-  tt*^M-.^+»-W 

65.  Find  the  value  of 

(a_  V6)(a-f  V6)(a—  V^l>)(a+  V^6)  when  a  =  2  and  6=3. 

66.  Solve  and  verify  \ 

I  a;  -f  2  y  —  m2  +  mw  -f  ™2» 

67.  Simplify T^_^!l  (       e      \ . 

68.  Show  that 

(m2^-?!2)?!"1— m     m2  —  n2  _      m2-mw 
w_1  —  »i_1  m3  +  w3  7i  —  m 

69.  Solve  and  test  the  solution  of 

^/5~x~^a  —  V5  a;  +     .  :  =  0 . 

V5a;  — a 

70.  Draw  the   graphs   of  2sc  +  3?/  =  7  and  4 <c  +  6 y  =  14, 

and  discuss  the  result. 

71.  Simplify  $  V 1  -  x  +  a  +  (»  -  a)  (1  -  x  +  a) ~*. 

3  Vm  +  n  +  2Vm-n 


72.   Rationalize  the  denominator  of 


3  Vm  +  n  —    V' 


m  — n 


73.  Solve    V2aj  +  5-f  V8aj-3=  V2o?  +  l,   and  test    the 
solution. 

74.  Simplify  [a^afW^]  •  [^ar^??]  + 

75.  Simplify 

V2+V^l  •  V2-V:=l '  V3  +  V^2  •  V3-V^2. 


CHAPTER   XXI 
QUADRATIC  EQUATIONS 

301.  A  quadratic  equation  is  an  equation  that,  in  its  simplest 
form,  contains  the  second  power,  but  no  higher  power,  of  the 
unknown  quantity. 

A  quadratic  equation  is  an  equation  of  the  second  degree 
(Art.  64). 

PURE  QUADRATIC  EQUATIONS 

302.  A  pure  quadratic  equation  is  an  equation  containing 
only  the  square  of  the  unknown  quantity.     Thus : 

ax*  =  a,  ^  =  16,  2/2  =  20  a\ 

A  pure  quadratic  equation  is  called  also  an  incomplete 
quadratic  equation. 

303.  Every  pure  quadratic  equation  may  be  reduced  to  the 
form 

x*  =  c. 

In  this  typical  form  we  note : 

(1)  The  coefficient  of  x2  is  unity. 

(2)  The  constant  term,  c,  may  represent  any  number,  posi- 
tive or  negative,  integral  or  fractional. 

Extracting  the  square  root  of  both  numbers  of  the  equation 

a?  =  c, 
we  obtain  x  =  ±  Vc7 

and  both  values,  +  Vc  and  —  Vc,  satisfy  the  given  equation. 


PURE  QUADRATIC  EQUATIONS  267 

The  omission  of  the  double  sign  before  the  square  root  of 
ie  left  member  has  no  effect  on  the  result,  no  root  being  lost 
by  the  omission.     For: 

(1)  4.  x  =  +-  Vc,  (3)  +aj  =  —  Vc,  and 

(2)  -z=-Vc,  (4)  -z=  +  Vc. 

From  (1)  and  (2),   x  —  Vc,  and  from  (3)  and  (4),  x—  —  Vtf. 

Clearly,  nothing  is  gained  or  lost  by  the  omission  of  the 
double  sign  before  the  square  root  of  the  left  member,  and  it 
is  not  customary  ,to  write  it. 

From  the  foregoing  we  have  the  method  for  solving  pure 
quadratic  equations : 

304.   Reduce  the  given  equation  to  the  form  x2  =  c. 
Extract  the  square  root  of  both  members,  observing  both  posi- 
tive and  negative  roots  of  the  nght  member. 

Illustrations : 

1.   Solve  9 a2- 7  =  2 a8 +  21. 

9x2-7  =  2a;2  +  21. 


Transposing, 

9ie2-2z2  = 

=  7  +  21. 

Uniting, 

7x2  = 

=  28. 

Dividing  by  7, 

X2  = 

=  4. 

Extracting  square  root, 

X  = 

:  ±  2.    Result. 

Verifying, 

9 

(±2)2-7  = 

:  2  (±2)2  +  21. 

9-4-7  = 

=  2.4  +  21. 

29  = 

=  29. 

2.   Solve       a  +  x_ 
a  —  x 

x  —2a 

x  +  2a 

Clearing  of  fractions, 

2aa 

+  3ax  +  x2  = 

-x*  +  Sax-2a*. 

Transposing, 

x2  +  x*  = 

-2a2-2a2. 

Uniting, 

2x*  = 

-4aa. 

Dividing  by  2, 

«2=: 
X  = 

-2a*. 

Extracting  square  root, 

±  V-2a2. 

Or, 

X- 

:  ±  a  V-2.     Result 

268 


QUADRATIC   EQUATIONS 


Solve: 

•1.     ^  =  49. 

2.  3ar»  =  75. 

3.  5a2-80  =  0. 

10.  5ar*  +  3 

11.  2^  +  7 


Exercise  101 

4.  4a2  =  9. 

5.  6a2 -18=0. 

6.  2a^==5. 
2x2  +  27. 
5x>  -  20. 


7.  3^  +  1  =  0. 

8.  ±ax2-l6a2  =  0. 

9.  7/  =  35a4. 


12.  3a2  +  18  -  <e  (x  +  1)  +  x  =  0. 

13.  aj(3»  +  2)-3  =  (z  +  1)2. 

14.  4s2  -  3*  +  2  =  (2*  -  1)(«  -  1). 

15.  (x  +  l)3  -{x-  l)3  =  26. 

16.  (3  +  x)  (x2  -  2)  =  3  +  x  (x2  -  2). 
(x  -  2)3  -  6  (aj  -  2)2  =  x  (x  -  4)  (a> 

a;2  -  5  1  1 


17. 
18. 


19. 


20. 


21. 


9). 


x2  -\-x  —  6      07+3 


X  +  1        X 


0. 


a  —  3     a  +  4 

or*  +  a;  —  1      x2  —  x  — 
x+2  ~x—2 

a^  +  a?-2  =  a32  +  a?-l 
<c2-a;  +  2      tf-x+i 


=  0. 


AFFECTED  QUADRATIC  EQUATIONS 

305.  An  affected  quadratic  equation  is  an  equation  con- 
taining both  the  square  and  the  first  power  of  the  unknown 
quantity,  but  no  higher  power. 

z2 +12  a  =  18.         3x2-2z-7  =  0.         ax2  +  bx  +  c  =  0. 

An  affected  quadratic  equation  is  called  also  a  complete 
quadratic  equation. 


AFFECTED   QUADRATIC   EQUATIONS  269 


Completing  the  Square 


306.  The  solution  of  an  affected  quadratic  equation  is  based 
upon  the  formation  of  a  perfect  trinomial  square.  In  the  per- 
fect trinomial  square, 

x2  -f-  2  ax  +  a2, 

we  note  (1)  The  coefficient  of  x2  is  unity. 

(2)  The  third  term,  a2,  is  the  square  of 

one  half  the  coefficient  ofx. 

That  is,  a2=f~\\ 

Similarly  in  x2  + 12 ax  +  36,         36  =  f~\\ 

lnx2-8ax  +  16a2,  16a2  =  f-  ^Y. 

307.  Given,  therefore,  an  expression  containing  the  first 
power  of  x,  and  the  second  power  of  x  with  the  coefficient  unity, 
we  may  form  a  perfect  trinomial  square  by  adding  to  the  ex- 
pression the  square  of  one  half  the  coefficient  ofx.     Thus,  given, 

x2  +  6  x,         x2  +  6  x  +  (3)2,  or  x2  -  6  x  +  9,      a  perfect  square. 

x2  -  18  x,       x2  -  18  x+  (  -9)2,     or  x2  -  18  x  +  81,     a  perfect  square. 
3  x2  +  4  x,       x2  +  (|)  x  +  (f  )2,      or  x2  - 1  x  +  f,      a  perfect  square. 

Oral  Drill 

With  each  expression  in  the  proper  form,  give  the  term 
necessary  to  make  a  perfect  trinomial  square. 

1.  a^  +  10a;.  5.  y2-S2y.  9.  3^  +  4^. 

2.  tf  +  Ux.  6.  ar>  +  7a;.  10.  5a^-8a;. 

3.  ^-16x.  •       7.  tf-Sx.  11.  2a^-3aj. 

4.  ^  +  24^.  8.  2/2+ll2/.  12.  6^-17  x. 

The  principle  of  completing  the  square  is  applied  to  the  solu- 
tion of  affected  quadratic  equations  in  the  following  : 


270  QUADRATIC  EQUATIONS 

Illustrations : 

1.  Solve  a2 +  12  a  =13. 

Adding  to  both  members  the  square  of  one  half  the  coefficient  of  x, 
x2  +  12  x  +  (6)2  =  13  +  36 
=  49. 
Extracting  square  root,  x+  6  =  ±  7. 

Whence,  x  =  ±  7  -  6. 

That  is,  x=  +  7-6orx=-7-6. 

From  which,  x  =  1  or  x  =  —  13.     Result. 

Both  values  of  x  are  roots  of  and  satisfy  the  given  equation. 
Verification : 

Ifx  =  l:      l2.+  12(l)  =  13,      1+12  =  13. 

Ifx=-13:      (-13)2  +  12(-13)=13,      169-156  =  13. 

2.  Solve  2z2-3a;-20  =  0. 

2x2-3x-20  =  0. 
Transposing,  2  x2  -  3  x  =  20. 

Dividing  by  2,  x2  —  |  x  =  10. 

Completing  the  square,    x2  —  f  x  +  (f  )2  =  10  +  ^ 

Extracting  square  root,  x  —  f  =  ±  l¥3-. 

Whence,  x  =  \6-,  or  -  J^. 

Or,  x  =  4,  or  —  f .     Result. 

Verification : 

If  X  =  4 :       2  (4)2  -  3  (4)  -  20  =  0.       32  -  12  -  20  =  0. 

Ifx=-f:       2(-f)2-3(-|) -20=0.  ^+Y__20=0. 

«     o  i       x*  —  x-\-l      q?  +  x  —  1     ., 

3.  Solve  J J— - — =1. 

x—2  x+2 

Clearing  of  fractions, 

(x2  -  x  +  l)(x  +  2)  -  (x2  +  x  -  l)(x  -  2)  =  (x  +  2)(x  -  2). 
x8  +  x2  -  x  +  2  -  x3  +  x2  +  3  x  -  2  =  x2  -  4. 
Whence,  x2  +  2  x  =  -  4. 

Completing  the  square,         x2  +  2  x  +  1  =  — 4  +  1 

=  -3. 
Extracting  square  root,  x  +  1  =  ±  V— 3. 

From  which,  x  =  —  1  ±  V—  3.    Result. 


AFFECTED   QUADRATIC  EQUATIONS  271 

Irrational  roots  result  when,  after  completing  the  square,  the  right 
member  of  an  equation  is  irrational. 

The  original  equation  will  be  satisfied  by  either  (—  1  +  V—  3)  or 

(-l-v^3). 

From  these  illustrations  we  may  state  the  general  process  for 
completing  the  square  in  the  solution  of  affected  quadratic 
equations: 

308.  Simplify  the  given  equation  and  reduce  to  the  form 
x2  -j-  bx  =  c. 

Add  to  each  member  the  square  of  one  half  the  coefficient  of  x. 

Extract  the  square  root  of  each  member,  and  solve  the  two 
resulting  simple  equations. 

Exercise  102 

Solve  and  verify : 

1.  ^-4a?==12.  11.   15x*  +  l±x=-S. 

2.  a?  +  2x=z35.  12.   14  a2 -  5  a;  =  24. 

3.  arJ-6a?  =  27.  13.   20<c2  +  aj  =  l. 

4.  a^  +  3a;  =  10.  14.   12ar*  +  23 o:  +  10  =  0. 

5.  x2-\-x  =  6Q.  15.   a2  +  4z  =  3. 

6.  2x2  —  Sx  =  2.  16.   x2-6x=6. 

7.  2>x*-lx  =  Q.  17.   x2-\-5x-S  =  0, 

8.  3^  +  17 a;+20  =  0.  18.   2ars-8a  =  l. 

9.  4:X2-5x  —  6  =  0.  19.   x2-4:X=-9. 
10.   6x2-x  =  2.                               20.   3a^-a;  +  l=0. 

21.  2x2-{-Sx=  -10. 

22.  (3a  +  l)2-(4«  +  l)(2a;-l)  =  ll. 

23.  (x4-l)-(x2  +  2)(x2-3)-(x  +  5)=0. 

24.  (2x  +  3)(x-2)-(3x-iy  =  x(x-3)+l. 

25.  (2 z  + 1) (3a -2)-(z  +  l) (2z -!)  =  (»  + l)(3a?-l). 


272  QUADRATIC   EQUATIONS 

26.  x2-2(x-l)  +  x(x-T)-(x-l)(x-2)  =  0. 

27.  x(x2 -x-3)  +  (x  +  3)(x  +  ll)  =  (x  —  1) (x2  +  1). 

28  2    't      3     -1.          34.    ^±±Z^«^±£±2. 

'  #  +  1      x  —  1                          ^  +  #  +  2      a^  +  x  — 3 

89.  «±!  +  i«4.                  35.    E±|  +  l  =  ^3. 

oj  +  3     a;     3  x  —  3  #  +  3 

30.  i±i  +  ^L-^l.          36.    -A- =6+1- 


05  +  1  a?  — 4      a? +  4             x'2  —  16 

31  a?  +  1   I  Ez^asl  37        -       I      X     ==a?2  +  5a;+8, 
'    x  —  1      a +  2        "  '    a  +  2     a  +  5      a2 +  7  a +  10* 

32  g-1  ,  g-r2^3  a+3     g-2  =  2(s-5) 

'    x  -2     ai-3     4*  x2        a2  +  l       x4  +  a^  ' 

33  ?*  +  l  ,3*  +  2  =  0  39     2^  +  1      3a?-l=5a?-2 
2x-l      'Sx-2       '  '     a?  +  l        x-1  "     a+2 

3z  +  4  2         x2  +  x-l 


40. 


^  +  2^  +  4     2  — aj        a^-8 


The  Quadratic  Formula 

309.   Every  affected  quadratic  equation  may  be  reduced  to 
the  form 

ax2  +  bx  +  c  —  0, 

in  which,  form  the  coefficients,  a,  b,  and  c,  represent  numbers, 
positive  or  negative,  integral  or  fractional. 


310.    Solving  this  affected  quadratic  equation  by  completing 

the  square,  we  have, 

ax2  +  bx  =  —  c. 

a  a 

a         \2a]  a      4a2 

=  62  -  4  ac 
4  a2 


AFFECTED  QUADRATIC  EQUATIONS  273 


x  + 

b 
2a 

X 

Vb2  -  4  ac 
4  a2 

b        V62  -  4  ac 
2a             2a 

_  -b±Vb2-,4ac 

2a 

This  value  of  x  from  the  general  equation,  ax?-{-  6a;  +  c  =  0, 
serves  as  a  formula  for  the  solution  of  affected  quadratic  equa- 
tions. The  formula  is  expressed  in  terms  of  the  coefficients  of 
the  given  general  equation,  and  by  substitution  of  particular 
values  for  a,  b,  and  c  from  a  given  equation  we  obtain  the  roots 
of  that  equation.  The  formula  is  the  most  practical  of  the 
many  methods  of  solution  and  it  should  be  memorized. 

To  obtain  the  solution  of  an  affected  quadratic  equation  by 
means  of  the  general  formula : 

311.  Transpose  all  terms  of  the  given  equation  to  the  left 
member,  and  reduce  to  the  form  ax2  +-  bx  +  c  =  0. 

In  the  formula  substitute  the  coefficient  of  the  given  x2  for  a, 
the  coefficient  of  the  given  x  for  b,  and  the  given  constant  for  c. 

Simplify  the  resulting  expression. 


Illustrations : 

1.   Solve  by  the  formula,  2  x2  +  5  x  =  12. 

Transposing,                  2  x2  +  5  x  - 

-  12  =  0. 

For  the  formula : 

0                     Then  in 
a  =  2, 

6  =  5, 

c  =  -  12. 

x  _  -  6  ±  V62  -  4  ac 
2a 

-(5)±V(5)2-4(2)(-12) 
2(2) 

_  -  5  ±  V25~+  96 

4 
_  -5  ±11 
4 

=  f ,  or  —  4.     Result. 

SOM.    EL.    ALG.  —  18 

274  QUADRATIC  EQUATIONS 

2.    Solve  by  the  formula,  x2  —  ±g. x  =  -§-. 
Transposing  and  clearing  of  fractions, 

3x2-10z-8  =  0. 
For  the  formula,        a  =  3,  b  =  —  10,  c  =  —  8. 


Substituting,    ^=-(-10)±V(-10)2-4(3)(-8). 

6 
x  ='4,  or  —  f .     Result. 

Exercise  103 

Solve  by  the  formula : 

1.  x?  -11  x  +  24  =  0. 

2.  o2-9#  =  22. 

3.  ^  +  9  #  +  14  =  0. 

4.  aj*-lla>=-28. 

5.  ^  +  9z  =  52. 

6.  a^-a;-72  =  0. 

7.  o*-21  z  =  46. 

8.  6o2-a  =  2. 

9.  2ar>+7a;  =  l5. 

10.  8x2  +  2x  =  3. 

11.  15^  +  44^  =  20. 

The  Solution  by  Factoring 

312.  Many  affected  quadratic  equations  may  be  solved  by 
an  application  of  factoring.  This  method  is  based  upon  the 
principle  that : 

313.  The  product  of  two  or  more  factors  is  zero  when  one  of 
the  factors  is  equal  to  zero. 


12. 

21  a*  +  29^-10  =  0. 

13. 

14^  +  53^  +  14  =  0. 

14. 

30  a2 -11  a  =  30. 

15. 

42  <&  —  m  so  =  -  20. 

16. 

a;2--u- #  +  2  =  0. 

17. 

^  +  -V3^  =  f 

18. 

x2--  =  2. 
6 

19. 

4     8. 

20. 

^_*?=o. 

30     3 

AFFECTED   QUADRATIC   EQUATIONS  275 

To  solve  an  affected  quadratic  equation  by  factoring : 

314.  Reduce  the  given  equation  to  the  general  quadratic  form, 
ax2  +  bx  -f-  c  =  0,  and  factor  the  resulting  trinomial. 

Assume  that  each  factor  in  turn  equals  zero,  and  solve  the  other 
factor  for  the  value  of  the  unknown  quantity. 

Illustrations : 

1.  Solve  by  factoring,  x2  +  6  x  =  7. 
Transposing,  x2  -f  6  x  —  7=0. 
Factoring,                                        (x  +  7)  (x  -  1)  =  0. 
From  which,                            x  +  7  =  0,  or  x  —  1  =  0. 
Solving,                                           x  =  -  7  and  x  =  1.     Result. 

2.  Solve  by  factoring,  2x2  +  1^x=%. 

Simplifying,  10  x2  +  11  x  -  6  =  0. 

Factoring,  (2  re  +  3)  (5  x  -  2)  =  0. 

Whence,  2  as  +  3  =  0  and  5  x  -  2  =  0. 
And  x  =  -f,  or  x  =  % .    Result. 

Exercise  104 

Solve  by  factoring: 

1.  a^-5a;  =  14.  10.   8a*-38a;  =  -35. 

2.  a?-8x  +  15  =  0.  11#   15  x2  -  77  a  + 10  =  0. 

3.  «2-3a;  =  4. 


4.  z2- 13  a +  12  =  0. 

5.  a^-12a;-13  =  0. 

6.  arJ-19<c  =  42. 


12.  12a?2-23a;  +  10  =  0. 

13.  x*-lf-x=*. 

14.  ^  +  ^+1=0. 


7.  2z2-9z  +  4  =  0.  1S    iB2_»==I. 

8.  4^-5^=6.  4     2 

9.  6^-17a;  =  -12.  16.   f a>  =  3»-ff. 


276  QUADRATIC   EQUATIONS 

LITERAL  AFFECTED  QUADRATIC  EQUATIONS 

315.    Any  one  of  the  three  methods  given  applies  readily  to 
literal  affected  quadratic  equations.    . 

Illustrations : 

1.    Solvec?(xi-l)=x(x  +  2c). 

Clearing  of  parentheses,  c2x2  —  c2  =  x2  +  2  ex. 

Transposing,  c2x2  —  x'1  —  2cx  =  c2. 

Uniting  coefficients  of  x2,  (c2  —  1)  x2  —  2  ex  =  c2. 

Dividing  by  (c2  -  1)  ,  x2  -  (j^)  *  =  j£j  ■ 

c2 


Completing  the  square,    x2  -  -^-  +  (  —^—r)   =  -^—-  +  — 

c2  —  1     \c2  —  1/       c2  —  1      (c 


2  -  l)2 

=  c2(c2-l)  +  c2 

(C2  -  l)2 

(c2-!)2' 


Extracting  square  root,  x -—r— -  =  ± 

c2  —  1 


±     * 


C2  _  1  -  C2  _  X 

From  which,  x  =  — ^—  orx= —  •     Result. 

c-  1  c  +  1 

2.  Solve *+»•-*£• 

Transposing  and  clearing  of  fractions, 

n2x2  +  mnx  -  2  m2  =  0. 
For  the  formula  :      a  =  n2,  b  =  ran,  c  =  —  2  m2. 


x_-mn±  V  (m»)2 - 4  (n2) (-2  m2) 
2(w2) 
_  —  mn  ±  V9  m2n2 

In2 
_  —  mn  ±  3  mn 
2n2 

=  »,  or-2-^.    Result. 


DISCUSSION  OF  AFFECTED  QUADRATIC   EQUATION      277 

Exercise  105 

Solve: 

1.  x2-2ax  =  3a\  8.  a?-2ax  +  a*  =  l. 

2.  x2  =  5  ax  —  6  a2.  9.  cV  +  c(m  —  ri)x  =  mn. 

3.  x2  +  ax  =  2a2.  10.  mx2—  (m  +  l)x  +  l  =  0. 

4.  3  aV  +  acx  —  2  c2.  11.  acx2  +  anx  =  cmx  +  mn. 

5.  IOcAb2- 21  c«a>  + 9  =  0.    12.  cV-c(a  +  l)«  +  a  =  (). 

6.  35c6x2  —  c3x  =  6.  13.  (2*+a)(*— c)=(aj— a)(»+c). 

7.  cc2  +  4a»  =  4a  +  l.  14.  (as  +  a -f- mf  =  (a  —  m)2. 

15.   «2-2aa;-26x  =  c2-a2-6(2a  +  6). 
1iS    a;-i-a  .      a         5  -„        1  1  _      1         1 

lb. 1 ; —  —  •  17. — • 

a         x  +  a      2  m  —  x     m     c—x     c 

ma    x-2a  .  x  +  2a     2x2  +  3a2 

lo. 1 -— "  •  • 

cc  +  3  a     x  —  3  a       0^  +  9  a2 

DISCUSSION  OF  THE  AFFECTED  QUADRATIC  EQUATION 

Character  of  the  Roots 

316.    If  the  two  roots  of  the  affected  quadratic  equation, 
ax2  +  bx  -+-  c  =  0,  be  denoted  by  t-j  and  r2,  we  have  (Art.  310), 


—  b  +  V&2  —  4  ac  —  6  —  V  &2  —  4  etc 

2  a  2a 

The  character  of  the  result  in  each  root  depends  directly 
upon  the  value  of  the  radical  expression  V&2  — 4ac,  for  the 
radicand,  b2  —  4  ac,  may  be  positive,  zero,  or  negative. 

(I)    When  b2  —  kac  is  positive. 

(a)  The  roots  are  real,  for  the  square  root  of  a  positive  quan- 
tity may  be  obtained  exactly  or  approximately. 

(6)  The  roots  are  unequal,  for  Vb2  —  4  ac  is  -f  in  rx  and  — 
in  r2. 

(c)  The  roots  are  rational  if  62  — 4ac  is  a  perfect  square, 
irrational  if  not.  ■ 


278  QUADRATIC   EQUATIONS 

Illustrations : 

(1)  In2sc2  +  lla;  +  12=0,  a=2,  6  =  11,  c=12.     62-4ac=121-96=25. 
Hence,  the  roots  of  2<c2+ll  x  +  12=0  are  real,  unequal,  and  rational. 

(2)  In5z2  +  lla  +  3  =  0,  a  =  5,  6  =  11,  c  =  3.     62-4ac=121-60=61. 
Hence,  the  roots  of  5  x2  +  11  x  +  3  =  0  are  real,  unequal,  and  irrational. 

(II)  When  b'2  —  4:ac  equals  0. 

(a)  The  roots  are  real,  and  ]  -™  -.  -,  b 

\(  _.  \  For  each  root  reduces  to . 

(6)  The  roots  are  equal.       J  2  a 

Illustration : 

In  4s2-20x+25=0,  a=4,  6  = -20,  c=25.    62-4ac=400-400=0. 

Hence,  the  roots  of  4  x'2  —  20  x  +  25  =  0  are  real  and  equal. 

(III)  When  b2  —  4  ac  is  negative. 

(a)   The  roots  are  imaginary,  for  the  square  root  of  a  nega- 
tive number  is  impossible. 

Illustration : 

In  3s2  +  2s +  2=0,  a  =  3,  6=2,  c  =  2.     62  -4ac  =  4  -24  = -20. 

Hence,  the  roots  of  3  x2  +  2  x  +  2  =  0  are  imaginary. 

317.    We  may,  therefore,  by  inspection  of  the  discriminant, 
b2  —  4  ac,  summarize  the  foregoing  conclusions  as  follows : 

I.   If  b2  —  4  ac  >  0,  the  roots  are  real  and  unequal. 

II.   If  b2  —  4  ac  =  0,  the  roots  are  real  and  equal. 

III.   If  b2  —  4  ac  <  0,  the  roots  are  imaginary. 

Illustrations : 

1.  Determine  the  character  of  the  roots  of  x*  —  5  x  =  6. 

aa-  5z-6  =  0.     a  =  l,  6  =  -5,  c  =  -6. 
62 _  4 ac  =  [(-5)2- 4  (1)(- 6)]  =  (25 +  24)  =  49. 
Therefore,  the  roots  are  real,  unequal,  and  rational.     (316, 1.) 

2.  Show  that  the  roots  ofar2  —  4sc+5=0  are  imaginary. 

62-4ac=[(-4)2-4(l)(5)]  =  (16-20)  =  -4. 
Hence,  the  roots  are  imaginary. 


DISCUSSION   OF  AFFECTED   QUADRATIC   EQUATION      279 

3.  Determine   the   value    of   m   for    which   the   roots   of 
4orJ  +  10a;  +  m=0  are  equal. 

By  Art.  316,  II,  the  discriminant  must  equal  0. 

Hence,         (10)2  -  4  (4)  (m)  =  0,  100  -  16  m  =  0,  m  =  ^. 

That  is,  the  roots  of  4  z2  +  10  *  +  ^  =  0  are  equal. 

4.  For  what  value  of  m  will  the  roots  of 

(m  +  l)a;2  +  (6m  +  2)a;  +  7m  +  4  =  0be  equal  ? 

The  discriminant  must  equal  0. 

a  =  (m  +  1),  6  =  (6m  +  2),c  =  (7m  +  4). 
Then       (6  m  +  2)2  -  4  (m  +  1)  (7  m  +  4)  =  8  m2  -  20  m  -  12. 
Solving,  8  m2  -  20  w  -  12  =  0,  m  =  3  and  -  J.    Result. 

By  substituting  these  values,  3  and  —  §,  in  the  given  equation,  two  dif- 
ferent equations  result,  both  of  which  have  equal  roots. 

Exercise  106 

By  inspection  of  the  discriminant,  determine  the  character 
of  the  roots  of : 

1.  a?  +  7x=&.  7.  2  x*  +  l  a  +  3  =  0. 

2.  a2-6a;  =  40.  8.  3  ar2- 5  x=  -  12. 

3.  tf2  +  5a-84  =  0.  9.  a?  +  10«  +  l  =  0. 

4.  ^-3^  +  54  =  0.  10.  3a^-12a;=-17. 

5.  x*  +  12x  =  -36.  11.  5ar>  +  40a;  =  l. 

6.  3a2  +  8a;  +  4  =  0.  12.  7  ar>  +  11  a?  +  12  =  0. 

Determine  the  values  of  m  for  which  the  two  roots  of  each 
of  the  following  equations  are  equal : 

13.  4x2  +  20a  +  ra  =  0.         16.   a2 -}- (m -f  5)  a;  +  5m+l  f=0. 

14.  9x*  +  mx  +  25  =  0.  17.   (m-f-l)a52-(m-2)a;-hl  =  0. 

15.  2m^-  30  #  —  15  =  0.     18.   2ma^+7ma!=^+5a;-5m. 

19.  (m  +  l)a;2  +  ma;=-9(a;  +  l). 

20.  m  —  7  +  mo2  —  m#  =  —  2  a;  —  x2. 


280  QUADRATIC   EQUATIONS 

Relation  of  the  Roots  and  Coefficients 

318.   If  the  roots  of 

ax2  +  bx  +  c  =  Oorx2  +  -x+-  =  0 
a        a 

are  obtained  by  Art.  310,  and  are  denoted  by  rx  and  r2  respec- 
tively, we  have 

2a  2a 

1*^  «aam^  „  _l  v       ~  b  +  V62  -  4  ac  -  b  -  V&2  -  4  ac 

By  addition,  n  +  r2  = ! 

2a 

Or,  ri+r2=_&. 

a 


By  multiplication,         nn  =  (- » +  V5' -  4«c)  (- 5  -  Vt° -4  ac) 

4  a2 
_  ft2  -  ft2  +  4  qc 
4  a2 

Or,  nr2=-- 

a 

Hence,  we  may  state : 

319.    In  any  affected  quadratic  equation  of  the  form 

a        a 


(a)  The  sum  of  the  roots  equals  the  coefficient  of  x  with  its  sign 
changed. 

(b)  The  product  of  the  roots  is  equal  to  the  constant  term. 

Illustrations : 

1.  Find  by  inspection  the  sum  and  the  product  of  the  roots 
of6x2  +  5x  =  6. 

Changing  to  the  required  form,  x2  +  f  x  —  1  =  0. 
From  Art.  319  : 

Sum  of  the  roots  =  —  f . 

Product  of  the  roots  =  —  1.    Result. 

2.  One  root  of  2  x2  +  5  x  =  12  is  —  -§.     Find  the  other  root. 

Transposing  and  dividing,  x2  +  f  x  —  6  =0. 

Dividing  the  product  of  the  roots,  -  6,  by  the  known  root,  -  f,  we 
have,  (-6)-(-f)=+(6x|)=4,     Result. 


DISCUSSION   OF  AFFECTED   QUADRATIC   EQUATION      281 

Exercise  107 

Find  by  inspection  the  sum  and  the  product  of  the  roots  of : 

1.  x>  +  9x  +  U  =  0.  5.   3ar>-10a;  +  3  =  0. 

2.  x2-21x  =  ±6.  6.   2ar>  +  3a  =  2. 

3.  2 a^+7 »«.!«.  7.  7^  +  9^-10  =  0. 

4.  6x2-x  =  12.  8.   15^  +  14^  +  3  =  0. 

9:   One  root  of  5  x2  —  26  x  +  5  =  0  is  5.    Find  the  other  root. 

10.  One  root  of  8  x2  =  15  x  +  2  is  —  -|-.     Find  the  other  root. 

11.  With    the    values    of    r,    and    r2   from   the    equation 
ax2  +  foe  +  c  =  0,  find  the  value  of 


rs—r 


1   .  1 


n  +  r2 


Also  the  value  of  —  + 


Formation  of  an  Affected  Quadratic   Equation    with 
Given  Roots 

b         c 
Consider  again  the  form  x2  +  -x  +  -  =  0.  (1) 

a        a 

If,  as  before,  i\  and  r2  denote  the  roots  of  this  equation,  we 
have  (Art.  319) : 

7*1  +  7*2= 

a 
Whence,      -  =  -  rs  -  r2  (2).  Also  (319),  -=  rxr2  (3) 

Ct  CL 

Substituting  in  (1)  the  values  found  in  (2)  and  (3), 

x2  +  (  —  rx  —  r2)x  +  rxr2  =  0. 

x2  —  rxx  —  r2a*  +  r^  =  0. 

(x2  —  r&)  —  (i\x  —  rfo)  =  0. 

x(x  —  r2)  —  rx  (x  —  r2)  =  0. 

(»-r1)(a;-r2)=0. 


282  QUADRATIC   EQUATIONS 


Therefore,  to  form  a  quadratic  equation  that  shall  have  any 
given  roots : 


320.    Subtract  each  root  from  x,  and  equate  the  product  of  the 
resulting  expressions  to  0. 

Illustrations : 

1.  Form  the  equation  whose  roots  shall  be  3  and  7. 

By  Art.  320,  (*  -  3)  (x  -  7)  =  0. 

Or,  x2  -  10  x  +  21  =  0.     Result. 

2.  Form  the  equation  whose  roots  shall  be  f  and  —  -|. 
By  Art.  320,  (z_f)(x_(_|))  =  o  ; 

(x-f)(x  +  §)  =  0; 
l5x2  +  x-6  =  Q.    Result. 

Exercise  108 

Form  the  equations  whose  roots  shall  be : 


1. 

2,5. 

5.    -2* 

8. 

a  —  1,  a  -f- 1. 

!      3 

9. 

a  —  1,  2  a. 

2. 

3,  -8. 

6     5         7 

6*  2'  ~8 

10. 

2±V=3. 

3. 

-4,-7. 

11. 

V2,  -1. 

4. 

3,-L 

7.  -?,-3. 

12. 

Va  +  1    Vtt— 1 

'      2 

3'      4 

2      '       2 

GRAPH  OF  A  QUADRATIC  EQUATION  IN  ONE  VARIABLE 

321.  The  graph  of  a  quadratic  equation  is  obtained  by 
application  of  the  principles  governing  the  graphs  of  linear 
equations.  The  given  equation  is  written  in  the  typical  form, 
ax2  +  bx  +  c  —  O,  and  the  left  member  is  equated  to  y.  By  as- 
suming successive  values  for  sc,  the  corresponding  values  of  y 
are  obtained  as  before. 


GRAPHS  OF  QUADRATIC  EQUATIONS 


GRAPHS  OF  QUADRATIC  EQUATIONS  HAVING  UNEQUAL  ROOTS 


283 


322.   Plot  the  graph  of 
a?  +  2x  =  8, 

Assume  y  =  x2  +  2  x  —  8. 
In  the  figure, 
If  x  =      3,  y  =      7.  P. 

X=      2,y  =      0.  Pi. 

x  =      1,  y  =  -  5.  P2. 

X  =      0,  y  =  -  8.  P8. 

x  =  -  1,  y  =  -  9.  P4. 

x=  -2,y=  -8.  P6. 

x  =  -  3,  y  =  -  5.  P6. 

x  =  -  4,  y  =      0.  P7. 

x  =  -  5,  y  =      7.  P8. 
etc. 


Bsag 


i 


The  curve  representing  the  equation,  ^  +  2^  —  8  =  0,  might 
be  indefinitely  extended  by  choosing  further  values  of  x.  In- 
termediate points  on  the  curve  may  be  obtained  by  assuming 
fractional  values  of  x  and  obtaining  corresponding  values  of  y. 

323.  The  lowest  point  of  the  graph  of  a  quadratic  equation 
in  one  variable  may,  in  general,  be  obtained  by  completing  the 
square. 

Thus,  x2  +  2x  +  1=8+  1;  (x  +  l)2  =  9;  (x+l)2-9  =  0. 
Now  (x  +  l)2  —  9  has  its  greatest  negative  value  when  x  =  —  1. 
Hence,  the  coordinates  of  the  lowest  point  of  the  curve  are  (—  1,  —  9). 

Plot  the  graph  of  2  x2  +  7  x  -  4  =  0. 

Let  y  =  2  x2  +  7  x  -  4.     Then  in  the  first  figure  on  p.  284 : 

(The  student  will  note  that,  in  the  figure,  the  scale  of  the  graph  is  so 
chosen  that  one  unit  of  division  on  the  axis  corresponds  to  two  units  from 
the  solutions.) 


284 


QUADRATIC  EQUATIONS 


If  x  =     2,  y  =  18. 
x  —     1,2/  =  5. 
x  =     0,  y  =  -  4. 
s=  —  1,  2/ =-9. 


P. 
Pi. 
P2 
Ps 


2 

[T 

P 

fi 

i  . 

1  . 

-1 

-it 

f 

X'                 ( 

)t x 

in: 

X 

t- 

T^ 

t. 

ll| 

fL 

J 

I 

B 

~Y" 

a;  =-2,  y  =-10. 
x=-S,y=-7. 
x=-4,y  =  0. 
x=-b,  y  =  11. 


P4. 
Ps- 
P6. 
P7. 


On  completing    the    square  the  lowest 
point  in  the  curve  is  (—  |,  -  -8^). 


Exercise  109 

Plot  the  graphs  of: 

1.  x>-§x  +  5z=0. 

2.  x2  +  6x  +  8  =  0. 

3.  a^  +  a;  — 12  =  0. 

4.  4  x2  —  5  x  =  0. 

5.  2^-5x  =  -3. 

6.  8a2  +  2z-3  =  0. 


GRAPHS  OF  QUADRATIC  EQUATIONS  HAVING  EQUAL  ROOTS 

324.   Plot  the  graph  of 


x>- 

-6x 

+  9 

=  0. 

Let    y  : 

=  x*- 

-6s 

+  9. 

the  figure 

Ifx  = 

0,2/ 

=  9. 

P. 

«  = 

1*9 

=  4. 

Pi. 

x  = 

2,2/ 

=  1. 

P2. 

X  = 

3,2/ 

=  0. 

P3. 

X  = 

4,  y 

=  1. 

P4. 

X  = 

5,  y 

=  4. 

Ps. 

X  = 

6,2/ 

=  9. 

P6. 

In 


« 


PV 


45 


Now,  ce2  —  6se  +  9  =  0,  may- 
be written    (x  —  3)2  =  0 ;   and 
x  in  this  equation  can  never  be  0.     Hence,  the  curve  cannot  cut  the  XX1 
axis,  but  is  tangent  to  it  at  (3,  0),  the  lowest  point  of  the  graph. 


GRAPHS  OF  QUADRATIC  EQUATIONS 


285 


GRAPHS  OF  QUADRATIC  EQUATIONS  HAVING  IMAGINARY  ROOTS 

325.    Plot  the  graph  of  x2  -  x  +  2  =  0. 

Let  y  =  x'2  —  x  +  2.     In  the  figure : 

x  =  3,  y  =  $. 
x  =  —l,  y  =  4. 
x  =  -  2,  y  =  8. 


lix  = 

0,  y  =  2. 

P. 

x  - 

1,  y  =  3. 

Pi. 

X  - 

2,  y  =  4. 

P2. 

x  = 

in 

Completing   the    square 
x2-x  +  2  =  0: 

X2 

-  x+2 

=  (z2- 

=  (x- 

-i)2  +  |. 

i  +  2 

p3. 
p4. 
p5. 


-  3,  y  =  14.     P6. 


Pif1 

EEEEEEEEK2EEEEEEE 

FTP 


And  this  expression  not 
being  0  or  negative  for 
any  value  of  x,  it  follows 
that  y  cannot  be  0  or  nega- 
tive ;  that  is,  the  graph 
cannot  touch  the  XX'  axis. 
Important  Conclusion. 
The  graph  of  any  quadratic 
equation   in    one   variable 

and  in  the  form  ax2  +  bx  +  c  =  0,  intersects  the  XX'  axis  at 
two  points  whose  abcissas  are  roots  of  the  equation,  provided 
that  the  given  equation  has  real  and  unequal  roots  ;  has  one 
point  in  the  XX'  axis  if  the  roots  are  equal ;  and  does  not 
cut  the  XX'  axis  if  the  roots  are  imaginary. 


Exercise  110 


Plot  the  graphs  of : 

1.  a2-4#  +  4  =  0. 

2.  4ar>-|-4:r  +  l=0. 


4.  x2  +  2x  +  S=0. 

5.  x?  +  x+6  =  0. 

6.  2x2-6x  +  ll  =  0. 


CHAPTER   XXII 

THE  QUADRATIC  FORM.    HIGHER  EQUATIONS. 
IRRATIONAL  EQUATIONS 

EQUATIONS  IN  THE  QUADRATIC  FORM 

326.  An  equation  in  the  quadratic  form  is  an  equation 
having  three  terms,  two  of  which  contain  the  unknown  num- 
ber ;  the  exponent  of  the  unknown  number  in  one  term  being 
twice  the  exponent  of  the  unknown  number  in  the  other  term. 
Thus: 

HIGHER  EQUATIONS  SOLVED  BY  QUADRATIC  METHODS 

327.  It  will  be  seen  at  once  that  many  equations  in  the 
quadratic  form  must  be  of  a  higher  degree  than  the  second. 
The  method  of  factoring  permits  the  solution  of  many  such 
equations,  and  is  generally  employed  in  elementary  algebra. 


328.  If  quadratic  factors  result  from  the  application  of 
factoring  to  higher  forms  of  equations,  they  may  ordinarily 
be  solved  by  completing  the  square  or  by  the  quadratic 
formula.  Such  factors  most  frequently  occur  in  connection 
with  binomial  equations. 


Illustrations : 

1.   Solve  x4  — 

13  x2  +  36  =  0. 

Factoring, 
And,     0  +  2)  (i 
"Whence, 

(z2 
c  -  2)  (x 

-  4)  («2  _  9)  = 
+  3)  (x  -  3)  = 
x  = 
286 

=  0. 

=  0. 

-2,2, 

3,  and  3.     Result. 


EQUATIONS  IN  THE  QUADRATIC  FORM  287 

2.  Solve  a?  +  7a?-8  =  0. 

Factoring,  (x3  +  8)  (x3  -  1)  =  0. 

And,     (as  +  2)  (x2  -  2x  +  4)  (x  -  1)  (x2  +  x  +  1)  =  0. 
By  inspection,  x  =  —  2  and  1. 

Solving,  x2-2x  +  4=0. 

x  =  1  ±  V^3. 
Solving,  x2  +  x  +  1  =  0. 

Therefore,  x  =  -  2,  1  ±  V^3,  1,  and  ~     +^  ~    ■    Result. 

3.  Solve  ^?  -  4a/^  =  12. 

Expressed  with  fractional  exponents  and  in  the  transposed  form,  we 
have, 

a;f  _4»i_  12  =  0. 

Factoring,  (x*  -  6)  (x^  +  2)  =  0. 

Whence,  x^  =  6,  or  x*  =  -  2. 

"From  which,  x  =  216,  or  —  8.    Result. 

When  tested  both  roots  prove  to  be  solutions. 

4.  Solve  x*  -  12  x~l  =  -  1. 

Jk     12  _      i 

With  positive  exponents,  x         7 —        * 

x* 
Clearing  of  fractions,  x^  —  12  =  —  x*» 

Transposing,  x^  +  x*  —  12  =  0. 

Factoring,  (x^  +  4)  (x*  -  3)  =  0. 

Whence,  x*  =  —  4,  or  x*  =  3. 

And,  x  s  256,  or  81. 

The  equation  is  satisfied  by  the  negative  value  of  v/256  and  by  the 
positive  value  of  VH. 

329.  For  convenience  of  solution  a  single  letter  may  be 
substituted  for  a  compound  expression  in  equations  in  the 
quadratic  form.  Care  should  be  taken  that  no  solution  is  lost 
in  the  final  substitutions. 


288       THE  QUADRATIC  FORM.     HIGHER   EQUATIONS 

Illustrations : 

1.    Solve  (a2  -  4)2  -  (a2  -  4)  =  20. 

Transposing,      (x2  -  4)2  -  (x2  -4)  -  20  =  0. 
Let        (x2  -  4)  =  y. 


Then, 

y2-y- 

20  =  0. 

Factoring, 

(y  -  5)  (y  + 

4)  =0- 

Whence, 

y  =  5,  or  —  4. 

If  y  =  5, 

If*/ =  -4, 

x2  -  4  =  5, 

X2  -  4  =  -  4, 

x2  =  9, 

x2  =  0, 

x  =  ±  3. 

x  =  0. 

Therefore,  x  =  ±  3  and  0 

Result. 

2.    Solve 


Let 


X  +  IV   ,   »  +  1 


X 

x  + 


\) 


+ 


=  y- 


x-l 

Then,  y2  +  y  -  6  =  0. 

Factoring,  (j/  +  3)  (y  -  2)  =  0- 
And,  y  =  —  3  and  2. 

If  2/  =  -  3, 


x+1 


x-l 
Hy  =  2, 

x  +  l. 
x-l" 


=  -3,    x  +  l  =  -3(x-l),        4x  =  2, 


r  Result. 


2,         x  +  1  =  2(x  -  1),  -  x  =  -  3,     x  =  3. 


3.    Solve  4a^  + 36  a;-2  =  25. 
Transposing,  4  x2  +  36  x~2  -  25  =  0. 


Then, 


4z2+!^_25 
x2 


Whence,  4  x4  +  36  -  25  x2  s=  0. 

Or,  4x4-25x2  +  36  =0. 

Factoring,  (x2  -  4)  (4  x2  -  9)  =  0. 

Or,  (x  +  2)  (x  -  2)  (2  x  +  3)  (2x  -  3)  =  0. 

And,  x  =  -  2,  2,  -  } ,  f.     Result. 


EQUATIONS   IN   THE    QUADRATIC   FORM 


289 


Exercise  111 


Solve : 

1.  a;4 -5ar>  + 4  =  0. 

2.  z4-8ar9  =  9. 

3.  a;-2- 3  a;-1  =  10. 

4.  x9  +  9a?  +  $  =  0. 

5.  a6-7ar*  =  8. 

6.  ar*  +  7a;-2  =  144. 

7.  »-1  +  ^  =  20. 

8.  2ar*  +  5ar*  +  2  =  0. 

9.  3  a  +  \/3  =  2. 


10.  or1  —  3ar*  =  4. 

11.  6aBr*  +  13«r*  +  6=aO. 

12.  8ar°-3ar2  =  -10. 

13.  2^  +  ^  =  7. 

14.  </?+ 3^-10=0. 


15.  Va?-5^  =  24. 

16.  a*  =  l. 

17.  (3x  +  l)(x2  +  x-2)  =  0. 

18.  2(«»  -  9)  =  3(»  -  3). 

19.  ^-8  +  3»(a;-2)  =  0. 

20.  3arJ  +  15a;  =  a<3arJ  +  17a;4-10). 

21.  2(ar}-8)+7arJ-17a;  +  6  =  0. 

22.  (a?  4-  3)2  -  5(x  +  3)  =  14. 

23.  (ar2  +  2)2  -  6(a^  +  2)  =  55. 

24.  (or*  +  a;)2  -  8(a?  +  x)  +  12  =  0. 

25.  (a  +  5)  -(aj  +  5)^  =6. 

26.  (x2  +  3a;  +  6)  -  2(ar>  +  3  a;  +  6)*  -  8  =  0. 

27.  a^  +  a;-f-2  =  7vV  +  a;  +  2-10. 

28.  (x  +  —  Y_3(W—  ^  +  2  =  0. 
ar>  +  2 


29. 


a^-2 
a^  +  2 


4-8-6^ 


30. 

SOM.    EL.    ALG 


?-2 


aT3T+°(a?T2j     =6 


19 


290       THE   QUADRATIC  FORM.     HIGHER  EQUATIONS 
EQUIVALENT  EQUATIONS  AND  THE  REJECTION  OF  ROOTS 

330.  If,  in  two  equations  involving  the  same  unknown 
number,  the  solutions  of  each  include  all  the  solutions  of  the 
other,  the  equations  are  equivalent. 

Thus,        3  x  X  2  =  x  +  6.     (1)  Whence        2  x  =  4.     (2) 

Equations  (1)  and  (2)  are  equivalent,  for  x  —  2  is  a  solution  j 
of  each. 

331.  In  the  solution  of  fractional  equations,  and  in  the 
solution  of  irrational  equations  also,  it  must  be  remembered 
that  the  processes  of  simplification  may  introduce  roots  that  | 
will  not  satisfy  the  given  equation.  That  is,  equivalent 
equations  do  not  always  follow  in  the  successive  steps  of  a  process. 

(a)  Processes  that  do  not  change  the  Roots  op  an  Equation  I 

6 


Given  the  equation  x  +  3 


x-2 


Clearing,  (x  +  3)  (x  -  2)  =  6. 

From  which,  x2  +  x  -  12  =  0. 

And,  x  =  3,  or  —  4. 

By  trial  both  3  and  —  4  are  found  to  be  roots  of  the  given  equation. 
Hence,  multiplying  by  x  —  2,  the  lowest  common  denominator,  intro- 
duced no  new  root. 

The  only  root  that  could  be  introduced  by  the  multiplier,  x  —  2,  would 
be  the  root  of  the  equation,  x  -  2  =  0,  or  x  =  2.  And  this  root  has  not 
been  introduced. 

In  general : 

332.    The  roots  of  a  fractional  equation  are  unchanged  if: 

(1)  the  given  fractions  having  a  common  denominator  are  com- 
bined, and 

(2)  the  equation  is  multiplied  by  the  loivest  common  multiple  of 
their  denominators. 


EQUIVALENT  EQUATIONS  291 

(b)  Processes  that  change  the  Roots  of  an  Equation 
Given  the  equation  x  +  5  =  0. 

Multiplying  by  as  - 1,     (se  -  1)  (x  +  5)  =  0. 

From  which,  x  =  —  5,  or  1. 

The  solution  x  =  1  fails  on  trial  to  satisfy  the  given  equation. 

In  general : 

333.  A  root  is  introduced  if  both  members  of  an  integral 
equation  are  multiplied  by  an  expression  involving  the  unknown 
number  in  the  equation. 

Given  the  equation  x  —  2  =  0. 

Transposing,  x  =  2. 

Squaring,  x2  =  4. 

Whence,  a2  -  4  =  0. 

Factoring,  (x  +  2)  (x  -  2)  =  0. 

Whence,  x  =  -  2,  or  2. 

Of  these  two  solutions  only  the  solution,  «  =  2,  satisfies  the  given 
equation,  the  solution,  x  =  —  2  being  introduced  by  the  process  of  squar- 
ing. 

In  general : 

334.  A  root  is  introduced  if  both  members  of  an  equation  are 
raised  to  the  same  power. 

The  Rejection  of  Roots. 

Since  it  is  clear  that  certain  processes  may  affect  the  solu- 
tions of  involved  types  of  equations,  we  make  the  following 
conclusion : 

335.  Before  accepting  all  the  solutions  of  a  given  equation  as 
roots  of  that  equation  it  is  necessary  that  each  solution  be  tested; 
and  all  solutions  not  satisfying  the  given  equation  must  be  rejected. 


292       THE   QUADRATIC   FORM.     HIGHER   EQUATIONS 
IRRATIONAL  EQUATIONS  INVOLVING  QUADRATIC  FORMS 

336.    Illustration: 


Solve  V2 x  +  7  —  V#—  5  =  VS. 

Squaring,  2  x  +  7  -  2  V(2x  +  7)(x  -  5)  +  x  -  5  =  x. 
Transposing,  —  2  V2  x2  —  3  a;  —  35  =  —  2  a;  —  2. 

Dividing  by  -  2,  V2  x2  -  3  x  -  35  =  x  +  1. 

Squaring,  2  x2  -  3  x  -  35  =  x2  +  2  35  +  1. 

From  which,  x2  -  5  x  -  36  =  0. 

And,  x  =  9,  or  —  4.      Result 

The  solution,   x  =  9,  satisfies  the  given  equation,  but  the  solution, 
x  =  —  4,  does  not  satisfy,  and  is  rejected. 

Note.     If  the  solution,  x  =  —  4,  is  tested  and,  in  extracting  the  square 
root  of  the  right  member  the  negative  value  of  the  root  is  taken,  we  have :  i 


V2(-4)+  7-V(-4)-6=  V-4. 
t/Zi  -  V^9  =  */-!. 

V^T-8  V^T  =  -2 
But  this  process  is  at  variance  with  the  accepted  condition  that  positive 
square  roots  are  to  be  consistently  taken  through  the  practice  of  ele- 
mentary algebra. 

Exercise  112 

Solve  and  test  the  solutions  of : 


1.  aj  —  l  =  V*  +  l.  4.    x  +  V2  =  Va  +  2  x  V2  +  4. 

2.  Vz2  +  «  +  l  =  V3a  +  4.         5.    V2(aj»-4)  =  (aj-2). 


3.   x  +  Va  +  l  =  19.  6.    ar2  +  a-3  =  Va4-r-2arJ+l0. 


7.    Vx-f-2-  Va-3  =  V 


.*• 


8.    (a:  +  3)(a>-l)  =  Va>(a8-7)  +  10  +  2a. 


Vg  +  l--v^-l=!i3a.4.2t 
Va  + 1  +  Vx  —  1 


FORMULAS  INVOLVING  QUADRATIC   EQUATIONS     293 
PHYSICAL  FORMULAS  INVOLVING  QUADRATIC  EQUATIONS 

Exercise  113 

1.  From  the  formula,  E  =  -—,  obtain  an  expression  for  v. 

2.  From  the  formula,  H=  .24<7V£,  obtain  an  expression  for  C. 

3.  Find  expressions  for  I  and  g  when  t  =  it -i/- . 

4.  Find  an  expression  for  t  from  the  formula  S  =  \gt2. 

Wv2 

5.  Given,  J57  =  -— — ;  find  an  expression  for  v. 

6.  Given  the  formulas,  V=gt,  and  S=^gt2.    Find  an  expres- 
sion for  the  value  of  t  in  terms  of  V  and  S. 

7.  Given  the  formula,  F  =  — - — .     Find  a  formula  for  t  in 

t  • 

terms  of  F,  m,  n,  and  r. 

8.  Given  the  formula,  t  =  v\j- .     Find  the  value  of  I  when 

Li.  v» 

9.  From  the  formula,  S  =  V0t  +  igt2,  obtain  the  value  of  t 
in  terms  of  V0,  g,  and  S. 

10.  Given  the   formulas,  E=Fs,  F=ma,  and  v=  V2as; 
obtain  a  formula  for  i?  in  terms  of  m  and  v. 

v2 

11.  If  a  =  — ,  and  t*=2  wr,  find  a  formula  for  a  in  terms  of 

r 

7r,  r,  and  £. 

12.  E  represents  the  energy  of  a  moving  body,  m  its  mass, 

7YIV 

and  v  its  velocity,  in  the  formula  E=  -—  •     What  is  the  rela- 

z 

tion  of  the  energies  of  two  bodies  if  one  has  twice  the  mass  but 
only  two  thirds  the  velocity  of  the  other? 


CHAPTER   XXIII 
SIMULTANEOUS  QUADRATIC  EQUATIONS.    PROBLEMS 

i 

337.  In  the  solution  of  simultaneous  quadratic  equations  we 
have  particular  methods  for  dealing  with  three  common  types ; 
but  no  general  method  for  all  possible  cases  can  be  given. 


SOLUTION  BY  SUBSTITUTION 


1.   Solve 


338.    When  one  equation  is  of  the  first  degree  and  the  other  q 
the  second  degree. 

'2x*-3xy-y*  =  l,  (1) 

_5x+y=3.  (2) 

From  (2),  #  y  =  S-6x.  (3) 

Substituting  in  (1),    2  x2  -  3  x  (3  -  5  x)  -  (3  -  5  a)2  =  1. 

2a;2  -  9x  +  15  x2  -  9  +  30x  -  25x2  =  1. 

8x2-21x  +  10  =  0.  (4) 

Factoring,  (8  x  -  5)  (x  -  2)  =  0. 

From  which,  x  =  f ,  or  2. 

Substituting  in  (3), 

Ifx  =  f,  y  =  3-5(|)  =  -f 

Ifx  =  2,  */ =  3 -5  (2)  =  -7. 
Hence,  the  corresponding  values  of  x  and  y  are 


, 


**»  •  1     Result. 


Corresponding  values  must  be  clearly  understood  as  to  mean- 
ing. From  (4)  in  the  above  solution  two  values  of  x  result. 
Each  value  of  x  is  substituted  in  (3),  and  two  values  for  y  result. 
Therefore,  we  associate  a  value  of  x  with  that  value  of  y  result- 
ing from  its  use  in  substitution. 

294 


SOLUTION   BY   COMPARISON   AND   FACTORING        295 


Exercise  114 
Solve : 

1.  x  +  2y  =  5,  6.    «2  +  an/  +  2/2  =  19, 
x*  +  3xy  =  7.  x  +  y  =  5. 

2.  x-3y=2,  7.   x*-3xy -y2  +  3  =  0, 
xy  +  y2  =  ti.  3x  +  2y  —  5  =  0. 

3.  2x  +  3y  =  12,  8.    3x  +  2y  =  5, 
xy-6  =  0.  2-xy-y2  =  0. 

4.  3x  +  2y  =  9,  9.    x?  +  y2  +  x  +y  =  6, 
xy  —  x  —  2.  x  —  4  y  =  6. 

5.  a;  +  22/  +  l  =  0,  10.    2  z  + 3  #  +  7  =  0, 
3ay-2/2  =  -4,  3aj*+2y*-a>  +  4y=:12. 


SOLUTION  BY  COMPARISON  AND  FACTORING 

339.    When  both  equations  are  of  the  second  degree  and  both 
are  homogeneous. 

Solve  (**-*-*  (*) 

13^-^  =  11.  (2) 

A  factorable  expression  in  x  and  y  results  if  the  constant  terms  are 
eliminated  from  (1)  and  (2)  by  comparison. 


Multiplying  (1)  by  11, 

22  x2  -  11  xy  =  110. 

(3) 

Multiplying  (2)  by  10, 

30  x2  -  10  y2  =  110. 

(4) 

Equating  left  members, 

30  x2  -10y2  =  22  x*  - 

-11 

xy. 

From  which, 

8x2  +  llxy  -  10y2  =  0. 

(5) 

Factoring  (5), 

(x  +  2y)(8x-5y)  =0. 

Therefore, 

,-f- 

(6) 

And, 

Sx 
V~  6  * 

• 

(7) 

296 


6              SIMULTANEOUS  QUADRATIC   EQUATIONS 

By  substitution  in  (1): 

Again  in  (1)  : 

From  (6)  :                  If  y  =  - 1, 

From  (7):             tfys**, 
5 

»*-*(-§)  =  10. 

2z2-x(^Wo. 

2x2  +  — =  10. 

2i 

2x2-^=10. 
5 

5a;2  =  20. 

2  x2  =  50. 

x2  =  4. 

x2  =  25. 

x  =±2. 
Hence,  in  (6), 

x—±  5. 
Hence,  in  (7), 

"=-ih-21(±2)=T1- 

*  =  if=!<:k6)=±& 

0         o 

Therefore,                    x  =  ±  2, 
y==Fl, 

x 

2/ 

=  ±5;1 

f     Result. 
=  ±8.J 

Note.    The  sign  T  is  the  result  of  a  subtraction  of  a  quantity  having 
the  sign  ±.     Thus,  —  (±  a)  =  —  (+  a)  or  —  (-  a)  =  —  a  or  +  a,  =  T  a. 


Solve : 

Exercise  115 

1.   x2-\-xy=6, 
xy—y2  =  l. 

6.    tf-xy  +  y2  =  %, 

x*  +  xy  +  y2  =  %. 

2.   x2  —  xy  —  4, 

7.   x2  —  3xy  +  y2  =  —  5, 

2ic2+a*/  -2?/2  =  -4. 

3.    x2-2xy  =  -8, 
y2-3xy  =  -9. 

8.   0^  +  ^  —  4^  =  18, 
a?2  —  a??/  —  3  y2  =  —  9. 

4.   x2  +  2«2/  =  9, 
3  xy  —  i/2  =  —  4. 

9.   10«2  +  7«2/-2/2  =  -ll, 
12x2-^-9xy-y2  =  -15. 

5.   ar2  +  5  xy  —  —  6, 

ar>  +  2/2  =  5. 

10.   2x2  +  4:xy  +  7y2-15  =  0, 
5^  +  3x2/-32/2-15  =  0. 

SOLUTION  OF  SYMMETRICAL  TYPES 


340.  When  the  given  equations  are  symmetrical  with  respect  to 
x  and  y ;  that  is,  when  x  and  y  may  be  interchanged  without 
changing  the  equations. 


SOLUTION   OF   SYxMMETRICAL  TYPES  297 


1.    Solve 


(x  +  y  =  7, 
[xy  =  10. 


(1) 
xy  =  10.  (2) 

Squaring  (1),  x2  +  2  xy  +  y2  =  49. 

Multiplying  (2)  by  4,  4  xy  =  40. 

By  subtraction,  x2  —  2  xy  +  y2  =  9. 

Extracting  square  root,  x  —  y  —  ±  3.  (3) 

Adding  (1),  x  +  y  =  7. 

2x=+3+7,  or-3  +  7. 

2  x  =  10,  or  4. 
x  =  5,  or2;l     Regult 
Subtracting  (1)  from  (3),  y  =  2,  or  5.  J 

2.    Solve  (^=12>  ft 

Multiply  (1)  by  2,  2  set/  =  24.  (3) 

Add  (2)  and  (3),  x2  +  2  xy  4  y2  =  49.  (4) 

Subtract  (3)  from  (2),         x2  -  2  xy  +  y2  =  1 .  (5) 

Extract  square  root  of  (4),  x  4  y  =  ±  7.  (6) 

Extract  square  root  of  (5),  x  —  y  =  ±  1.  (7) 

Four  pairs  of  equations  result  in  (6)  and  (7),  viz. : 
x  +  y=7  x  +  y=     7  x  +  y  =  -l  x  +  y  =  -7 

x  —  y  =  1  x  -  y  =—  1  x-  y  =     1  x  —  y  =  —  1 

x=4  x=3  x=-3  x  =  —  4 

y  =3  y=     4  y  =-4  y  =-3 

Hence,  the  corresponding  values  for  x  and  y  are 
x  =  ±4,         y=±3,i 
x  =  ±3;        y=±4.j 


Result. 


Solve : 


Exercise  116 


1.  a  +  y  =  6,  3.   ^  +  2/2  =  5, 
ojy  =  6.  as  +  y  =  3. 

2.  x  +  y  =  5,  4.    a#— 3=0, 
zy-4  =  0.  a^4-2/2  =  10. 


298  SIMULTANEOUS   QUADRATIC   EQUATIONS 

5.  x2  +  xy  +  y2  =  7,  7.   x2  +  y2=5, 
x  +  y  =  3.         i  (x  +  y)2  =  9. 

6.  x  +  y-5  =  0,  8.   o2  +  3a?/  +  2/2  =  59, 
x2  -xy  +  y2  =  7.  x2  -f  o?y  +  y2  =  39. 

SOLUTIONS  OP  MISCELLANEOUS  TYPES 

341.  Systems  of  simultaneous  quadratic  equations  not  con- 
forming to  the  three  types  already  considered  are  readily 
recognized,  and  the  student  will  gradually  gain  the  experience 
necessary  to  properly  solve  such  systems.  No  general  method 
for  these  types  can  be  given. 

It  is  frequently  possible  to  obtain  solutions  of  those  sys- 
tems in  which  a  given  equation  is  of  a  degree  higher  than 
the  second,  derived  equations  of  the  second  degree  resulting 
from  divisions  and  multiplications. 

.    Illustrations : 

l.   Solve  j^  +  3*  +  3,  =  28,  (1) 

lajy-6  =  0.  (2) 

Transposing  and  multiplying  (2)  by  2,  2  xy  =  12.  (3) 

Adding  (1)  and  (3),      x2  +  2  xy  +  y2  +  3  x  +  3  y  =  40. 

Whence,  (x  +  y)*  +  3  (x  +  y)  -  40  =  0. 

Factoring,  (x  +  y  +  8)  (x  +  y  —  5)  =  0. 

Hence,  x  +  y  =—  8,  or  x  +  y  =  5. 

Combining  each  of  these  results  with  (2),  we  have  two  systems : 

(a)  x  +  y  =  -  8,  (6)     x  +  y  =  5, 

xy  -  6  =  0.  xy  -  6  =  0. 

The  two  systems  (a)  and  (&)  may  be  readily  solved  by  the  principle 

of  Art.  340. 

*.   Solve  mf**-*  ■         I 

[x  —  y  =  l.  (*) 

From  (1),  xiy*  +  5  xy  -  14  =  0, 

(xy  +  7  j  (xy  -  2)  =  0. 


SOLUTIONS  OF  MISCELLANEOUS  TYPES 


299 


Whence,  xy  =  2,  or  xy  =  —  7. 

Combining  each  of  these  derived  equations  with  (2),  we  have, 
(a)  xy  =  2,  (6)  xy=-l, 

x  —  y  =  \.  x  —  y  =  l. 

Solving  these  two  systems,  we  find 

from  (a),  x  =  2,  or  —  1, 

y  =  1,  or  -  2  ; 

from  (&),  _  lj-3  V33   ^ 

*-  2 

_l±3v3 
y  2 


3.    Solve  J 


fl  +  l  =  ? 
1      1  =  3 

a    y    2 


Squaring  (2), 
Dividing  (1)  by  (2), 
Subtracting  (4)  from  (3), 
Whence, 


x2     xy     y2     4 


xy     y* 


_3  =6 
xy     4' 

JL  =  1 

xy     2 


Subtracting  (5)  from  (4) ,      —  -  —  +  -o  =  1 . 
x2     scy     y2     4 


x     y         2 


Extracting  square  root  of  (6) , 

x 

Combining  (2)  and  (7),  we  obtain,      x  =  *'  or  J  ;1 

y  =  2,  or  1.  J 


Result. 


4.   Solve 


*«  +  y*  =  82, 


se+y=4 

Let  x  =  w  +  v, 

and  y  =  u  —  v. 

Then  in  (1),  (u  +  t>)4  +  («  -  O4  =  82. 


(1) 
(2) 
(3) 
(4) 

(5) 
(6) 
(7) 


(1) 
(2) 


800  SIMULTANEOUS   QUADRATIC  EQUATIONS 

Or,  m4  +  4m3h    6  u2v2  +  4  uvs  +     v4  ' 

+  ul  —  4  w3t?  +    6  rt'V2  —  4  2fy3  +     w4 


2  w4  +12  ztV2               +  2 1?4  =  82 

Whence,  w4  +  6  w2v2  +  v4  =  41. 

Substituting  in  (2),  w  +  v  +  w  ~  v  =  4. 

w  =  2. 

Substituting  (5)  in  (4),  (2)4  +  6  (2)V  +  vi  =  41. 

Or,  tf  +  24 1>2  _  25  =  0. 

Whence,  v  =  ±  1,  or  ±5  V-  1. 

Therefore,  if  u  =  2  and  t>  ==±  J,  And,  if  u  =  2  and  v  =  ±5  y/~—  1, 

x  =  w-fv  =  2±l=3orl,  x=u+v=2±5  V^T, 

y  —  u  —  v  =  2  +  l=lor3.  2/  =  w-<y  =  2=F5  V—  1. 

Hence,  in  brief  form,    x  —  3,  1,  or  2  ±  5  V—  1;1      Resuit 
y  =  1,  3,  or  2  T5  V-1.1 

r(a-j0  +  (*-y)*  =  12,  (1) 


5.    Solve  . 

(z  +  2/)-(z  +  2,)*  =  2.  (2) 

Considering  (a;  +  yy  and  (x  —  y)  *  as  unknown  quantities  and  factoring : 
From  (1),  (x  -  y)  +  (x  -  */)*  -  12  =  0. 

(x  -  yy  =  3,  and  (x  -  */)*  =  -  4.  (3) 

From  (2),  (x  +  y)  -  (x  -  y)^  -  2  =  0. 

(x  +  y)^  =  2,  and  (x  +  yy  =  -l.  (4) 

The  negative  roots,  being  extraneous,  are  rejected.     Then, 


From  (3), 

(x  -  yy  =  3. 

From  (4), 

(x  +  y)$  =  2. 

Whence, 

x-y  =  9. 

(5) 

And, 

x  +  y  =  4. 

(6) 

From  (5) 

and 

(6), 

2  x  -  13,  and  2  y  = 

-5. 

Therefore, 

x  =  ^  and      y  = 

5 
1' 

Result. 

SOLUTIONS  OF   MISCELLANEOUS  TYPES  301 

Exercise  117 

Solve : 

1.  a?  +  y2  =  13,  16.   ar5  -  ^  -  98  =  0, 
2x  =  3y.  2-x  +  y  =  0. 

2.  x  +  y  =  4,  17.   x2-\-xy-i-y2  =  91, 
x2  —  2xy  +  3y  —  x  =  3.  x  +  Vxy  +  y  =  13. 

3.  x  +  2/  +  l=%y,  18.   x2  +  y2-\-x  +  y  =  l±, 
5x-y-xy  =  l.  x2_y2  +  x_y=zxid. 

4.  ^  +  2/2  -f  0:2/  =  39,  19    X2  +  Xz  +  Z2  =  30y 
x-y  =  3.  x*-xz+*  =  l&. 

5.  x  +  y  =  2,  2Q    tT5  +  2/5  =  244, 
12-x*y*  =  ±xy.  x  +  y  =  ±. 

6.  PV-»  +  ^  21.   ,-,-12  =  0, 

7.  a?  +  y  =  3a>y-l,  ■ 

^  +  /  =  8^2-3^-3.     22«  ^-^2/  +  2/2  =  3, 

8.  xy  +  x-y  =  b, 
fcy(a  -  2/)  =  6- 

9.  ^2  +  2/24-^  +  2/  =  2, 
a*/-2  =  0. 

10.  IB2  — 42/2  =  a;  +  2y, 

x  +  4  ?/  =  7. 

11.  3*4-^=^+10, 
x—y=xy+2. 

12.  x4+ary  +  2/4  =  19, 
ar2  —  a*/  4-  2/2  =  7- 

13.  x4  +  2/4-17=0, 
a;4-y--3  =  0. 

14.  (z-?/)2-a;2?/2-5  =  0, 

15    3a;2+2«?/-2?/2=14(a;-2/), 
2a?  +  xy-3y2  =  7(x-y). 


x  —  xy+y  =  l. 

23. 

it     12     6  y 

.-i-l-o. 

o, 

24. 

x2     y2     16 

25. 

1      1_7 

x*     f     $' 
1_1_1 
x     y     2' 

1      1      1 

+  ^  =  o, 

26. 

x  +  2/     »  — y 
2     3_12 
x     y     xy' 

+  5       ' 

302 


SIMULTANEOUS   QUADRATIC   EQUATIONS 


GRAPHS   OF  QUADRATIC  EQUATIONS  IN  TWO  VARIABLES 

TYPE  FORMS  OF  EQUATIONS  AND  CORRESPONDING  GRAPHS 


(I)  Type  Form 


7 

3J£Be 

^•jT    ^ 

-£          tn 

2               -\ 

ZZZZt                         J 

)r„± .x 

-±- — 5 

r                1 

t                      V                              ^E 

S^          -.^5 

_5ta;~*2B 

jegff 

K 

jr2  +/  =  c. 

Illustration : 
Plot  the  graph  of 

x2  +  y2  =  25. 

From  the  equation  we  have 


If 


y  =  ±  V25  -  P- 


jc=0,  y=  ±' V25,  or  5.  P. 

«s=X,  y=±V24,  or  ±4.89+.  Pi. 
jc=2,  y=±  V2l",  or  ±4.06+.  P2. 
a; =3,  y=  ±  Vie,  or  ±4.  P3. 
»=4,  y=±V9,  or  ±3.     etc. 

Plotting  these  points,  we  ob* 
tain  a  circle  as  the  graph  of  the 
equation,  x2  +  y2  =  25. 

It  will  be  seen  that  the 
coordinates  of  any  point  (x,  y) 
are  legs  of  a  right  triangle 
whose  hypotenuse  is  the  dis- 
tance from  the  origin  to  the 
point  (x,  y).  That  is,  for  any 
point  on  the  curve  we  have 
(see  figure), 

^  +  2/2=52, 
or,  x2  +  y2  =  25. 

In  general,  therefore: 

342.    Tfie  graph  of  any  equation  in  two  variables  in  the  form 
x2  +  y2  =  c  is  a  circle. 


GRAPHS  OF  QUADRATIC  EQUATIONS 


303 


(II)  Type  Form  •••  f  =  ax  +  c. 

Illustration : 
Plot  the  graph  of 

If 
x=0,  y  =  ±  V8,  or  ±2.8+.   P2. 

x=  l,y  =  ±V&,  or  ±3.4+.  P3. 

x=-l,  y  =  ±V4,  or  ±2.      Pl 

35= -2,  y=0;    etc.  P. 

(Note  the  enlarged  scale.) 

Plotting  these  points,  we  ob- 
tain a  parabola  as  the  graph 
of  the  equation,  y2  =  4  x  +  8. 


¥ 

^' 

*'% 

IS           «3 
X    P 

6^" 

2L 

y 

£ 

x           ir x 

PI             o 

V 

\ 

V 

S^vp 

1   ^s 

s5 

^N 

V 

It  will  be  seen  that  if  x  is  less  than  —  2,  y  is  imaginary ; 
hence,  no  point  in  the  curve  lies  to  the  left  of  P. 
From  this  type  form  and  graph  we  have,  in  general : 

343.    The  graph  of  any  equation  in  two  variables  in  the  form 
y2  =  ax  +  c  is  a  parabola. 

(Ill)  Type  Form  ••• 
ax2  +  by2  =  c. 

Illustration : 

Plot  the  graph  of 

±x2  +  9y2  =  36. 

If  y  =  0,  x2  =  9,  x  =  ±3.    A. 

x  =  0,  y2  =  4,  y  =  ±2.   B. 

Hence,  x=  +  3  and  —  3,  the 
points  where  the  graph  cuts 
XX  ;  and, 

y  =  +  2  and  —  2,  the  points 
where  the  graph  cuts  YY'. 


9       zn 

g        B     P 

^ j^ 

-7                 V 

+-     t                     \ 

xi : o i * 

A^                                       JA 

\                                        Z 

-S«                        ^ 

^._         -^ 

B     B  p 

Y^ 

304 


SIMULTANEOUS  QUADRATIC  EQUATIONS 


For  any  other  points  let  x  =  ±  1. 

Then,        9  y2  =  32,  y  =  ±  f  i/%  ttf  f  =s±  1 .9+. 

Hence,  for  P,  Pif  P2,  and  P8,  we  have  (1,  1.9),  (1,  - 1.9),  (- 1,  1.9), 
(—  1,  —  1.9)  respectively. 

Plotting  these  points,  we  obtain  an  ellipse  as  the  graph  of  the  equation 
4x3  +  9y2  =  36. 

By  assuming  values  sufficiently  greater  or  less  than  those  by 
which  the  points  above  are  obtained,  it  can  be  shown  that  no 
points  in  the  graph  can  lie  to  the  right  or  teft,  above  or  below, 
the  intersections  with  the  axes  of  reference. 


From  this  type  form  and  graph  we  have,  in  general : 

344.    The  graph  of  any  equation  in  two  variables  in  the  form 
ax*  -\-by2  =  c  is  an  ellipse. 

(IV)  Type  Forms  •  •  •  ax2  —  by2  =  c  and  xy  =  c. 


Illustrations : 

1.   Plot  the  graph  of 

x2-4y2  =  l. 
From  the  given  equation 


X 

s.                                  ^^** 

^£                 P  +  ' 

r                   -Sta                        -«£_ 

Hi 5I--A2 x 

/-    Q       *s 

>'                  s^ 

+'   £           _    U   ss 

*~ 

Y*~ 

Iix  =  ±l,y2  =  0,y=0. 
x  =  ±2,y*=%,y=±iy/3, 
or  ±.86+. 

For  the  values  of  x  we  plot 
A,  (1,  0),  and  5,  (-1,  0). 

AlsoP,  (2,  .86);  Pu  (-2,  .86); 
P2,(-2,-.86);P„(2,-.86). 

With  these  points  we  obtain 
an  hyperbola  as  the  graph  of 
the  equation,  x2  —  4  y2  =  1. 

It  will  be  found  by  trial  that  any  value  of  x  between  -f- 1  and 
—  1  gives  an  imaginary  value  for  y,  hence  no  part  of  the  curve 
can  lie  between  the  points  A  and  B. 


GRAPHS  OF  QUADRATIC  EQUATIONS 


305 


2.   Plot  the  graph  of 

If  x  =  -4,  •••    0    •••  +  4,  etc., 
y  =  _l,  •••  ±qo  ».  +  l,  etc. 

Plotting  the  points,  (—4,  —  1), 
etc.,  we  obtain  an  hyperbola 
whose  branches  lie  in  the  angles 
FOX  and  VOX*. 

From  the  two  cases : 

345.  TJie  graph  of  any 
equation  in  two  variables  in 
the  form  ax*  —  by2  —  c,  or  in  the  form  xy  =  c,  is  an  hyperbola. 

Exercise  118 


=  ||||||      *i 

till 

lllilll 

HH-Iy-  1 1 1 1 1 1  - 


Plot  the  graph  of : 

1.  x2  +  y2  =  16. 

2.  s2  +  y2  =  49. 


3.   y2  =  4a?4-4. 


5.  <c2-32/2  =  2. 

6.  xy  —  10. 


SOLUTION  OF  SIMULTANEOUS  QUADRATIC  EQUATIONS  BY  GRAPHS 

m 


/ 


,B 

=53E 


\ 


Illustrations : 

1.   Given#2  +  2/2  =  25, 
Sx-2y  =  6. 

From  the  intersections  of 
the  graphs  of  the  given 
equations  obtain  the  roots, 
and  check  the  results  by 
solving  the  equations. 

In  the  figure  the  graph  of 
x2  -f-  y2  =  25  is  a  circle,  and 
the  graph  of  3x  —  2y  =  6 
is  a  straight  line. 


SOM.    EL.    ALG. — 20 


306 


SIMULTANEOUS   QUADRATIC   EQUATIONS 


Y     ptm 

S* 

^ d>^ 

ii-zfiiiifc::::::: 

i£^l&^ 

J! 

-*>— ^PF-s- 

E^S 

o, 


By  measurement  of  the  graphs  we  find  the  coordinates  of 
P%  and  P2  to  be  (4,  3)  and  (—  -J-|,  —  -f-f ).  Solving  the  equations, 
we  obtain  x  —  4,  y  =  3 ;  or  x  =  —  -}-|,  y  =  —  ^|.  The  accuracy 
of  this  and  of  subsequent  cases  may  be  increased  by  plotting 
on  a  larger  scale. 

2.    Given 
x2  +  y2  +  9x  +  U 
y2=4:X  +  16. 
From  the  intersections 
of     the    graphs    of     the 
given     equations     obtain 
the     roots,     and      check 
the  results  by  solving  the 
equations. 

In  the  figure  the  graph 
of  x2  +  y2  +  9  x  +  14  =  0 
is  a  circle  and  the  graph  of 
2/2  =  4#  +  16isa  parabola. 
By  measurement  of  the 
graphs  we  find  the  coordinates  of  Px  and  P2  to  be  (—  3,  2)  and 
(—3,-2)  respectively,  and  these  coordinates  correspond  to 
the  real  roots  obtained  from  the  solution.  For  the  solution  of 
the  system  gives  x  =  —  3,  y  =  ±2,  or  x  =  —  10,  y  =  ±  V—  24. 
We  cannot  find  a  point  corresponding  to  the  imaginary  root, 
and  we  make  the  important  conclusion  that : 

346.  Since  there  can  be  no  point  having  one  or  both  coordinates 
imaginary,  the  graphs  of  two  equations  can  have  no  intersections 
corresponding  to  imaginary  roots. 

3.   Given  x2  +  4  y2  =  17, 
xy  =  2. 

From  the  intersections  of  the  graphs  of  the  given  equations 
obtain  the  roots,  and  check  the  results  by  solving  the  equations. 


GRAPHS  OF  QUADRATIC  EQUATIONS 


307 


Iii  the  figure  the  graph  of  a?2  +  4  ?/2  =  17  is  an  ellipse,  and 
the  graph  of  xy  =  2  is  an 
hyperbola. 

By  measurement  of  the 
graphs  we  find  the  co- 
ordinates of  Plt  P2,  P3, 
and  P4tobe'(4,  £),  (1,2), 
(-4,  -h),  (-1,  -2) 
respectively.  The  solu- 
tion of  the  equations  gives 
x=  ±4,  y=  ±i,  or  x=  ±1, 
|=  ±2. 

It  will  be  noted  that  this 
last  case  is  the  first  in 
which  both  equations  are 
homogeneous  and  of  the  second  degree,  and  that  the  graphs 
serve  to  emphasize  once  more  the  importance  that  attaches  to 
the  association  of  corresponding  values  of  x  and  y  in  the 

solution  of  a  system  of 
simultaneous  quadratic 
equations. 

4.    Given 

2a,-2  +  5y2  =125, 
5  x  +  2  y  =  10. 

From  the  intersections 
of  the  graphs  of  the  given 
equations  obtain  the  roots, 
and  check  the  results  by 
solving  the  equations. 

In  the  figure  the  graph 
of     2^+22/2  =  125     is 
an    ellipse,   and  the   graph   of  5  x  +  2  y  =  10   is   a    straight 
line. 


308 


SIMULTANEOUS   QUADRATIC    EQUATIONS 


Y 

"^w 

■\ 

^<  ""  m  "™  ^  <  P 

z    ~    qL 

2                §\       T 

^                 J-^Sr 

x: o s x 

^ 

I                        4                s 

V            -,£ 

^      ^ 

v-*».— tf^        

SI 

By  measurement  of  the  graphs  we  find  the  coordinates  of 
Pj  and  1\  to  be  (0,  5)  and  (3.7+,  4.4-)  respectively.      The 

solution  of  the  equations 
gives     x  =  0,    y  =  5,    or 
3  =  3.75+,  ?/ =4.38+. 
5.    Given, 

a;2  +  f  =  100, 
3a; +  4?/ =  50. 
From  the  intersections 
of  the  graphs  of  the  given 
equations  obtain  the  roots, 
and  check  the  results  by 
solving  the  equations. 

In  the  figure  the  graph 
of  x2  +  y2  =  100  is  a  circle, 
and  the  graph  of  3  x  +  4  y 
=  50  is  a  straight  line. 
Solving  the  system,  we  find  that  the  equation  resulting  from 
substitution  gives  equal  roots  for  y.  Hence,  for  x  we  find  but 
one  value,  the  solution  of  the  system  being  x  =  6,  y  =  8.  It 
will  be  seen  from  the '  graph  that  the  circle  and  straight  line 
have  one  point  only  in  common ;  that  is,  the  line  is  tangent  to 
the  circle.  The  coordinates  of  the  point  of  tangency  are  found 
to  be  (6,  8)  or  P.     In  general : 

347.  If,  in  the  solution  of  a  system  of  equations,  a  derived 
equation  has  equal  roots,  the  graphs  of  the  equations  are  tangent 
to  eadi  other. 

It  will  be  found  to  assist  the  student  in  the  exercise  follow- 
ing if  the  type  forms  of  equations  and  the  corresponding 
graphs  are  remembered. 

1 .  ax  +  by  —  c,  The  Straight  Line. 

2.  jr2  +/  =  c,  The  Circle. 

3.  /2  =  ax  4-  c,  The  Parabola. 


GRAPHS  OF  QUADRATIC  EQUATIONS       309 

4.  ax2  +  by2  =  c,  The  Ellipse. 

5.  ax2-b/=c,ov\  The  Hyperbola. 

6.  xy  —  c,  J 

Exercise  119 

Plot  the  graphs  of  the  following,  and  check  the  roots  deter- 
mined by  a  solution  of  each  system  : 

1.  x2+y2  =  25,  5.    x2  +  y2  +  x  +  y  =  18, 
x  —  y=l.  xy  +  x  +  y  =  11. 

2.  ^+2/2=74,  6.    r  =  4a  +  8, 


xy  =  35. 

«  4-  y  —  6  =  0. 

3. 

a;  —  y  =  6. 

7.   9 x2-  16y2  =  144, 
^4-^  =  36. 

4. 

a?  +  y  =  12, 

a*/ =  32. 

8.   0^+4^  =  4, 
x2  +  tf '■  =  17. 

9. 

a2 

a2 

4-2/2  = 
-xy 

=  32, 

+  2/2 

=  28. 

10.  Solve  the  system,  x  +  y  =  0  and  a^/  =  4,  and  determine 
if  the  solution,  alone  will  show  that  the  graphs  intersect,  or 
do  not  intersect. 

11.  How  do  the  graphs  of  xy  =  4  and  xl  —  y2  — 16  differ  in 
their  positions  relative  to  the  axes  of  reference  ? 

12.  Can  you  describe  the  position  of  the  graph  of  the  equa- 
tion, x2  +  y2  —  2  x  =  0,  without  plotting  it  ? 

13.  How  do  the  graphs  of  the  equations,  x2  -f  2y2  =  32  and 
4  #2  +  y2  =  16,  differ  in  their  relative  positions  ? 

14.  Plot  the  graphs  of  x2  —  xy  4-  y>  =  28  and  x  —  y  =  0,  and 
show  that  their  intersections  check  the  roots  found  by  solution. 

15.  In  how  many  points  may  the  graphs  of  x2  +  y2  =  25  and 
a,-2  4-  2  y2  =  64  intersect?     Prove  your  answer  by  plotting. 


810  SIMULTANEOUS  QUADRATIC  EQUATIONS 

16.  Show  the  positions  of  the  graphs  of  x2  +  y2.  =  100  and 
x2  +  4?/2  =  100,  and  determine  the  number  of  real  roots. 

17.  Show  by  solution  that  y2  —  4  x  =  12  and  x2  +  y2  -f  3  x  =  0 
can  have  but  one  point  in  common,  and  prove  your  answer  by 
a  graph  of  the  system. 


PROBLEMS  PRODUCING  QUADRATIC  EQUATIONS 

348.  In  the  solution  of  a  problem  from  which  a  quadratic 
equation  or  equations  result  we  retain  only  the  solution  that 
satisfies  the  given  conditions.  As  a  rule  negative  results  will 
not  ordinarily  satisfy  the  conditions  even  if  they  satisfy  the 
equations. 

Illustrations : 

1.  If  6  times  the  number  of  laborers  in  a  field  are  increased 
by  the  square  of  the  number  at  work,  there  will  be  in  all  55 
men.     How  many  laborers  are  there  in  the  field  ? 

Let  x  t=  the  number  of  laborers  in  the  field. 
From  the  given  conditions, 

6cc  +  «2  =  55. 
From  which,  x2  +  6  x  —  55  =  0. 

Solving,  x  as  5,  or  x  =  —  11. 

Clearly  the  positive  result  only  is  retained.    Hence,  5  laborers.     Result. 

2.  A  company  of  boys  bought  a  boat,  agreeing  to  pay  for  it 
the  sum  of  $  60.  Three  of  the  boys  failed  to  pay  as  agreed, 
so  each  of  the  others  was  compelled  to  pay  $1  more  than  he 
had  promised.     How  many  boys  actually  paid  for  the  boat? 

Let  x  =  the  number  of  boys  actually  paying  for  the  boat. 

x  +  3  =  the  number  of  boys  first  agreeing  to  share  its  cost. 

—  =  the  number  of  dollars  paid  by  each  boy. 

=  the  number  of  dollars  each  had  expected  to  pay. 


x  +  S 


PROBLEMS  PRODUCING  QUADRATIC  EQUATIONS     311 

Then,  ■  S>— S»-«i 

x      x  +  3 

From  which,  x2  +  3  x  -  180  =  0, 

and  x  =  12,  or  -  15. 

That  is,  12  boys  actually  shared  in  the  cost  of  the  boat. 

Many  of  the  problems  in  the  following  exercise  must  be 
stated  by  the  use  of  two  unknown  quantities,  and  simulta- 
neous quadratic  equations  will  result.  But  as  far  as  pos- 
sible the  student  should  attempt  to  state  most  of  the  earlier 
examples  by  means  of  one  unknown  number  only. 


Exercise  120 

1.  Find   those  two   consecutive  odd   numbers  the  sum  of 
whose  squares  is  290. 

2.  Find  a  number  that,  added  to  7  times  its  reciprocal, 
equals  8. 

3.  Two  factors  of  48  are  such  that  one  exceeds  the  other 
by  2.     What  are  the  factors  ? 

4.  The  sum   of  two  numbers  is  10,  and  the  sum  of  their 
squares  is  52.     Find  the  numbers. 

5.  Find  two  numbers  such  that  their  sum  and  the  difference 
of  their  squares  are  each  13. 

6.  Find  the  two  factors  of  600  whose  sum  is  49. 


7.  Find  two  numbers  whose  sum  is  11,  and  whose  product 
is  17  less  than  15  times  their  difference. 

8.  The  sum  of  the  cubes  of  two  numbers  is  126,  and  the 
sum  of  the  two  equals  6.     Find  the  numbers. 

9.  Find  two  numbers  the  sum  of  whose  squares  is  130  and 
whose  product  is  63. 


312  SIMULTANEOUS   QUADRATIC   EQUATIONS 

10.  How  many  yards  of  picture  molding  will  be  required 
for  a  room  whose  ceiling  area  is  1200  square  feet,  the  diagonal 
of  the  ceiling  being  50  feet  ? 

11.  One  of  two  numbers  exceeds  30  by  as  much  as  the  other 
number  is  less  than  30,  and  the  product  of  the  numbers  is  875. 
Ifind  them. 

12.  The  sum  of  the  squares  of  two  numbers  equals  13  times 
the  smaller  number,  and  the  sum  of  the  numbers  is  10.  What 
are  the  numbers  ? 

13.  If  twice  the  product  of  the  ages  of  two  children  is  added 
to  the  sum  of  their  ages,  the  result  is  13  years.  One  child  is 
3  years  older  than  the  other.     Find  the  age  of  each. 

14.  The  diagonal  of  a  rectangle  is  100  feet,  and  the  longer 
side  is  80  feet.     Find  the  area  of  the  rectangle. 

15.  Find  three  consecutive  numbers  such  that  the  sum  of 
their  squares  shall  be  194. 

16.  The  simple  interest  on  $600  for  a  certain  number  of 
years  and  at  a  certain  rate  is  $  120.  If  the  time  were  two 
years  shorter  and  the  rate  2  %  more,  the  interest  would  be 
$108.     Find  the  time  and  the  rate  of  interest. 

17.  The  sum  of  the  squares  of  two  numbers  is  2  a2  +  2,  and 
the  sum  of  the  numbers  is  2  a.     What  are  the  numbers  ? 

18.  The  combined  capacity  of  two  cubical  tanks  is  637  cubic 
feet,  and  an  edge  of  the  one  added  to  an  edge  of  the  other 
equals  13  feet.  Find  the  length  of  a  diagonal  on  any  one  face 
of  each  cube. 

19.  The  product  of  two  numbers  is  15  greater  than  5  times 
the  larger  number,  and  is  6  less  than  16  times  the  smaller 
number.     Find  the  numbers. 

20.  If  a  number  of  two  digits  is  multiplied  by  the  tens' 
digit,  the  product  is  96;  and  if  the  number  is  multiplied  by 
the  units'  digit,  the  product  is  64.     Find  the  number. 


PROBLEMS   PRODUCING   QUADRATIC   EQUATIONS     313 

21.  Divide  15  into  two  parts  such  that  their  product  shall 
equal  10  times  their  difference. 

22.  The  difference  of  two  numbers  is  1,  and  the  sum  of  the 
numbers  plus  the  product  is  19.     Find  the  numbers. 

23.  If  the  sum  of  two  numbers  is  multiplied  by  the  less, 
the  product  is  5 ;  and  if  the  difference  of  the  numbers  is  multi- 
plied by  the  greater,  the  product  is  12.     Find  the  numbers. 

24.  A  garden  40  feet  long  and  28  feet  wide  has  around  it  a 
path  of  uniform  width.  If  the  area  of  the  path  is  960  square 
feet,  what  is  its  width  ? 

25.  A  dealer  would  have  received  $  2  more  for  each  sheep 
in  a  drove  if  he  had  sold  6  less  for  $  240.  How  many  were 
there  in  the  drove,  and  at  what  price  was  each  sold  ? 

26.  In  a  number  of  two  digits  the  units'  digit  is  3  times 
the  tens'  digit,  and  if  the  number  is  multiplied  by  the  sum  of 
the  digits,  the  product  is  '208.     Find  the  number. 

27.  A  bicyclist  starts  on  a  12-mile  trip,  intending  to  arrive  at 
a  certain  time.  After  going  3  miles  he  is  delayed  15  minutes 
and  he  finds  he  must  travel  the  remainder  of  the  journey  at  a 
rate  3  miles  an  hour  faster  in  order  to  arrive  at  his  destination 
on  time.     Find  his  original  rate  of  speed. 

28.  The  sum  of  the  squares  of  the  two  digits  of  a  number 
is  13,  and  if  the  square  of  the  units'  digit  is  subtracted  from 
the  square  of  the  tens'  digit  and  the  remainder  is  divided 
by  the  sum  of  the  digits,  the  quotient  is  1.     Find  the  number. 

29.  The  difference  between  the  numerator  and  the  denomina- 
tor of  a  certain  improper  fraction  is  2,  and  if  both  terms  of 
the  fraction  are  increased  by  3,  the  value  of  the  fraction  will 
be  decreased  by  -£$.     Find  the  fraction. 

30.  From  the  formula,  t  =  vxj-9  find  the  length  of  a  pen- 

es' 
dulum  that  vibrates  once  a  second  at  a  point  where  g  =  32.16 
feet. 


314  SIMULTANEOUS   QUADRATIC   EQUATIONS 

31.  A  body  falls  through  a  space  of  3216  feet  at  a  point  where 
g  equals' 32.16  feet.  From  the  formula,  S  =  \  gt2,  determine 
the  number  of  seconds  required  for  the  fall. 

32.  If  an  automobile  traveled  3  miles  an  hour  faster,  it  would 
require  2  hours  less  time  in  which  to  cover  a  distance  of  120 
miles.  What  is  the  present  rate  of  the  automobile  in  miles  per 
hour  ? 

33.  A  certain  floor  having  an  area  of  50  square  feet  can  be 
covered  with  360  rectangular  tiles  of  a  certain  size ;  but  if  the 
masons  use  a  tile  1  inch  longer  and  1  inch  wider,  the  floor  can 
be  covered  with  240  tiles.     Find  the  sizes  of  the  different  tiles. 

34.  One  leg  of  a  right  triangle  exceeds  the  other  leg  by  2 
feet,  and  the  length  of  the  hypotenuse  is  10  feet.  Find  the 
length  of  the  legs  of  the  triangle. 

35.  168  feet  of  fence  inclose  a  rectangular  plot  of  land,  and 
the  area  inclosed  is  1440  square  feet.  Find  the  dimensions  of 
the  field. 

36.  The  sum  of  the  squares  of  two  numbers  is  increased  by 
the  sum  of  the  numbers,  and  the  result  is  18.  The  difference  of 
the  squares  of  the  numbers  is  increased  by  the  difference  of  the 
numbers,  and  the  result  is  6.    Find  the  numbers. 

37.  If  the  difference  of  the  squares  of  two  numbers  is  divided 
by  the  smaller  number,  the  remainder  is  4  and  the  quotient  4.  If 
the  difference  of  the  squares  of  the  numbers  is  divided  by  the 
greater  number,  the  remainder  is  3  and  the  quotient  3.  What 
are  the  numbers  ? 

38.  If  the  length  of  a  certain  rectangle  is  increased  by  2 
feet,  and  the  width  is  decreased  by  1  foot,  the  area  of  the  rect- 
angle will  be  unchanged.  The  area  of  the  rectangle  is  the 
same  as  the  area  of  a  square  whose  side  is  3  feet  greater  than 
one  side  of  the  rectangle.  What  are  the  dimensions  of  the 
rectangle  ? 


CHAPTER   XXIV 

RATIO.     PROPORTION.     VARIATION. 

RATIO 

349.  If  a  and  b  are  the  measures  of  two  magnitudes  of  the 
same  kind,  then  the  quotient  of  a  divided  by  b  is  the  ratio  of 
a  to  b. 

Ratios  are  expressed  in  the  fractional  form,  -,  or  with  the 

b 

colon,  a :  b.     Each  form  is  read  "  a  is  to  b." 

350.  In  the  ratio,  a :  b,  the  first  term,  a,  is  the  antecedent, 
and  the  second  term,  b,  is  the  consequent. 

THE  PROPERTIES  OF  RATIOS 

351.  The  properties  of  ratios  are  the  properties  of  fractions; 

for  the  ratios,  -,  m:n,  (#,+  y):(x  —  y),  etc.,  are  fractions. 
b 

(a)  The  Multiplication  and  the  Division  of  the 
Terms  of  Ratios 

352.  The  value  of  a  ratio  is  unchanged  if  both  its  terms  are 
multiplied  or  divided  by  the  same  number. 

353.  A  ratio  is  multiplied  if  its  antecedent  is  multiplied,  or  if 
Us  consequent  is  divided,  by  a  given  number. 

354.  A  ratio  is  divided  if  its  antecedent  is  divided,  or  if  its 
consequent  is  multiplied,  by  a  given  number. 

315 


316  RATIO.     PROPORTION.     VARIATION 


(6)  Increasing  or  Decreasing  the  Terms  of  a  Ratio 

355.  If  a,  b,  and  x  are  positive,  and  a  is  less  than  b,  the  ratio 
a :  b  is  increased  when  x  is  added  to  both  a  and  b. 

For  a  +  x     a  =  xQ*-a) 

'  b  +  x      b      b(b  +  x) 

And,  since  a  <  b,  the  resulting  fraction  is  positive  and  the 
given  ratio,  - ,  is  increased  accordingly. 

356.  If  a,  b,  and  x  are  positive,  and  a  is  greater  than  b,  the 
ratio  a:b  is  decreased  when  x  is  added  to  both  a  and  b. 

For  the  resulting  fraction  in  Art.  355  is  negative  when  a  >  b. 

357.  An  inverse  ratio  is  a  ratio  obtained  by  interchanging 
the  antecedent  and  the  consequent. 

Thus,  the  inverse  ratio  of  m  :  n  is  the  ratio  n  ;  m. 

358.  A  compound  ratio  is  a  ratio  obtained  by  taking  the 
product  of  the  corresponding  terms  of  two  or  more  ratios. 

Thus,  mx  :  ny  is  a  ratio  compounded  from  the  ratios,  m  :  n  and  x :  y. 

359.  A  duplicate  ratio  is  a  ratio  formed  by  compounding  a 
given  ratio  with  itself. 

Thus :  a2  :  62  is  the  duplicate  ratio  of  a  i  b. 

In  like  manner,  az :  63  is  the  triplicate  ratio  of  a  :  b. 


Exercise  121 

1.  Write  the  inverse  ratio  of  a :  x ;   of  m  :  n ;  of  7  :  12 ;   of 
3x:5x;  of  (2a  +  l)  :  (2a-  1);  of  (a?  +  xy  +  y2)  :  (x2-xy  +  y2), 

2.  Arrange   in    order   of   magnitude   the   ratios   2:5,  3:7, 
4 :  9,  5 :  8,  10  :  17,  12  :  19,  21 :  27,  32 :  39,  and  40 :  51. 

3.  Compound  the  ratios  3  :  7  and  10 :  17. 

4.  Find  the  ratio  compounded  of  3 :  8,  4 : 9,  and  6 :  11. 


PROPORTION  317 

5.  Compound  the  ratios  (x2  —  9) :  (ic3  +  8)  and  (x+2) :  (x—3). 

6.  What  is  the  ratio  compounded  from  the  duplicate  of 
2 : 3  and  the  triplicate  of  3 : 2  ? 

7.  Find  the  value  of  the  ratio  (x  +  6)  :  (x2  +  7  x  +  6). 

8.  Two  numbers  are  in  the  ratio  of  4:7,  but  if  3  is  added 
to  each  number,  the  sums  will  be  in  the  ratio  of  5 : 8.  Find 
the  numbers. 

9.  The  ratio  of  a  father's  age  to  his  son's  is  16  :  3,  and  the 
father  is  39  years  older  than  the  son.     Find  the  age  of  each. 

10.  In  a  certain  factory  5  men  and  4  boys  receive  the  same 
amount  for  a  day's  work  as  would  be  paid  if  3  men  and  12 
boys  were  engaged  for  the  same  time.  What  is  the  ratio  of 
the  wages  paid  the  men  and  the  boys  individually? 

PROPORTION 

360.  A  proportion  is  an  equation  whose  members  are  equal 
ratios. 

Thus,  the  four  numbers,  a,  6,  c,  and  d,  are  in  proportion  if  -  =  -. 

b     d 

361.  A  proportion  may  be  written  in  three  ways : 

(1)     2-£.  (2)     a:b  =  c:d.  (3)     a:b::C:d. 

b     d 

Each  form  is  read  "  a  is  to  b  as  c  is  to  d."  We  understand  the  mean- 
ing of  a  proportion  to  be  that  the  quotient  of  a  ■*■  b  is  the  same  in  value 
as  the  quotient  of  c  -=-  d. 

362.  The  extremes  of  a  proportion  are  the  first  and  fourth 
terms. 

363.  The  means  of  a  proportion  are  the  second  and  third 
terms. 


318  RATIO.     PROPORTION.     VARIATION 

364.  The  antecedents  of  a  proportion  are  the  first  and  third 
terms,  and  the  consequents  the  second  and  fourth  terms. 

In  the  proportion  a:b  =  c:d 

a  and  d  are  the  extremes,  a  and  c  are  the  antecedents, 

b  and  c  are  the  means  ;  b  and  d  are  the  consequents. 

365.  If  the  means  of  a  proportion  are  equal,  either  mean  is 
a  mean  proportional  between  the  first  and  fourth  terms. 

Thus :  in  a  :  b  =  b  :  c,  b  is  a  mean  proportional  between  a  and  c. 

366.  The  last  term  of  a  proportion  whose  second  and  third 
terms  are  equal  is  a  third  proportional  to  the  other  two  terms. 

Thus,  in  a:b  =  b  :c,  cisa  third  proportional  to  a  and  b. 

367.  A  fourth  proportional  to  three  numbers  is  the  fourth 
term  of  a  proportion  whose  first  three  terms  are  the  three  given 
numbers  taken  in  order. 

Thus,  in  a  :  b  =  c  :  d,  diss,  fourth  proportional  to  a,  6,  and  c. 

368.  If,  in  a  series  of  equal  ratios,  each  consequent  is  the 
same  as  the  next  antecedent,  the  ratios  are  said  to  be  in  con- 
tinued proportion. 

Thus  :  a  :b  =  b:c  =  c:d=d:e  =  etc. 

369.  In  the  treatment  of  proportions  certain  relations  are 

conveniently  discussed   if  we  recall  that  -  =  -=?*.     Whence 
J  b     d 

a  =  br,  and  c  =  dr.     Substitutions  of  these  values  for  a  and  c 
will  be  of  frequent  service  in  practice. 

370.  In  order  that  four  quantities,  a,  b,  c,  and  d,  may  be 
in  proportion,  a  and  b  must  be  of  the  same  kind,  and  c  and  d 
of  the  same  kind.  However,  c  and  d  need  not  be  of  the  same 
kind  as  a  and  b. 


PROPERTIES   OF   PROPORTION  319 

PROPERTIES  OF  PROPORTION 

371 .    Given  a:b  =  c:d.     Tlien  ad  =  be. 


Proof : 


b~ d 
Multiplying  by  bd,  ad  =  be. 


That  is: 

In  any  proportion,  the  product  of  the  means  equals  the  product 
of  the  extremes. 

372.    Givena:b  =  c:d.     Then  a  =  — ,  b  =  — ,  etc. 

d  c 


From  the 

equation  ad  = 

be 

we  obtain  by  division 

„      be 

d 

_bc 

—      » 
a 

o  =  — ,      c  = 
c 

ad 
b 

That  is : 

Either  extreme  of  a  proportion  equals  the  product  of  the  means 
divided  by  the  other  extreme;  and  either  mean  of  a  proportion 
equals  the  product  of  the  extremes  divided  by  the  other  mean. 

373.    Given  a:b  =  b:c.     Then  b  =  -Vac. 


Proof  : 

a :  b  =  b  :  c. 

By  Art.  371, 

b2  =  ac. 

Extracting  square  root, 

b  =  y/ac. 

That  is: 

The  mean  proportional  between  two  numbers  is  equal  to  the 
square  root  of  their  product. 

374.    Given  ad  =  be.   Then  a :  b  =  c :  d. 

Proof :  ad  =  bc. 

Dividing  by  bd,  9l  =  9:. 

b      d 

Or,  a:b  =  c:d. 


320  RATIO.     PROPORTION.     VARIATION 

That  is : 

If  the  product  of  two  numbers  is  equal  to  the  product  of  two 
other  numbers,  one  pair  may  be  made  the  extremes,  and  the  other 
pair  the  means,  of  a  proportion. 

In  like  manner  we  may  obtain  a  proportion,  a:c=b:d,  etc. 


375.    Given 

a 

6  = 

: c : d.     Then  b:a  =  d:c. 

Proof : 

a  _c 
b~ d 

Then, 

b            d 

Whence, 

b_d 
a      c 

Or, 

b:a  =  d:c.     . 

That  is : 

If  four  numbers  are  in  proportion,  they  are  in  proportion  by 
inversion. 

376.  Given  a:b  =  c:d.     Then  a:c  =  b:d. 

Proof:  °l  =  c-. 

b     d 

Multiplying  by  5,  «&  =  &£. 

c  be     cd 

Whence,  ^  =  -. 

c     d 

Or,  a  :  c  =  b  :  d. 

That  is : 

If  four  7iumbers  are  in  proportion,  they  are  in  proportion  by 
alternation. 

In  applying  alternation  all  four  quantities  considered  must 
be  like  in  kind. 

377.  Given  a:b  =  c:d.     Then  a+b:b  =  c +d:d. 

Proof:  2«'5. 

b     d 


PROPERTIES  OF  PROPORTION  321 


Adding  1  to  both  members,    -  +  1  =  -  +  1. 
b  d 


Whence, 

a  +  V  _c  +  d 
b             d 

Or, 

a  +  b  :b  =  c  +  d:d. 

That  is : 

If  four  numbers  are  in  proportion;  they  are  in  proportion  by 
composition. 

378.    Given  a:b  =  c:d.     Then  a  —  b:b  =  c  —  d:d. 


Proof : 

a  _  c 
b~  d 

Subtracting  1, 

|-  1  =  ^-1. 
b            d 

Whence, 

a  —  b      c  —  d 
b             d 

Or, 

a  —  b.b  =  c  —  d:d. 

That  is : 

If  four  numbers  are  in  proportion,  they  are  in  proportion  by 
division. 

379.    Given  a:b  =  c:d.     Then  a-\-b:a—b  =  c  +  d:c  —  d. 

Proof:  a±b^c±d,  (1) 

b  d 

\„a  a  —  b     c  —  d  /ON 

And,  —  =  — •  (2) 

Dividing  (1)  by  (2),  £±|  =  c-±^,- 

a  —  b     c  —  a 

Or,  a  +  b  :a  —  b  =  c  +  d:c  —  d. 

That  is: 

If  four  numbers  are  in  proportion,  they  are  in  proportion  by 
composition  and  division. 

80M.    EL.    ALG. 21 


322  RATIO.    PROPORTION.     VARIATION 

380.  Given  a:b  =  c:d.     Then  an  :bn  =  cn  :  dn. 

Proof:  Sfil 

6     d 

Raising  both  members  to  the  nth  power, 

9-  —  SH. 

bn  ~  dn ' 

Or,  an:bn=  cn:  dn. 

That  is : 

Like  powers  of  the  terms  of  a  proportion  are  in  proportion. 

381.  Given  a:b  =  c:d  =  e:f=  •••.    T7iew  (a  +  c  -+-  e  +  •••)  : 
(b  +  d  +  f+.>.)=a:b. 

Proof:     Since  ®  =  -?  =  *  =  ..., 

b     d    f         ' 

then,  -  =  r,  -  =  r,  -  =  r,  etc. 

6  d  / 

Whence,  a  =  &r,  c  =  dr,  e  =^fr,  etc. 

Adding,  a  -f  c  +  e  +  •••  =  br  +  dr  +fr  +  •••. 

Whence,  a  +  c  +  e  +  •••  =  (&  +  d  +  /  +  •••)»*. 

And,  q  +  c  +  e+-  =  r=g. 

Or,  (0  +  c+e  +  -)  :(&  +  <*  +/+•••)  =  a:&. 

That  is : 

Jn  a  series  of  equal  ratios,  the  sum  of  the  antecedents  is  to  the 
sum  of  the  consequents  as  any  antecedent  is  to  its  consequent. 


382.    Given  a 

:&  = 

=  b 

c. 

Then  a  :c—a2:  b2. 

Proof :    Since 

a     b 
6"c' 

it  follows  that, 

« v  b  _a     a 
b     c~b     b 

Whence, 

a_a2 
c     &2' 

Or, 

a  :  c  =  a2  :  b2. 

APPLICATIONS   OF  THE   PROPERTIES  OF  PROPORTION     323 

That  is : 

If  three  numbers  are  in  continued  proportion,  the  first  is  to  the 
third  as  the  square  of  the  first  is  to  the  square  of  the  second. 


383.    Given  a:b  =  b:c  =  c:d.     Then  a:d  =  a8:bs. 
Proof :     Since 


it  follows  that,  «x^x-  =  ?x?xa 


b~ 

_b_ 
c 

c 
=  ~d' 

c      d 

=  <** 
b 

a 
b 

d~ 

~b*' 

a 

d  = 

=  a* 

b* 

b     c     d     b     b     b 
Whence, 

Or, 
That  is : 

If  four  numbers  are  in  continued  proportion,  the  first  is  to  the 
fourth  as  the  cube  of  the  first  is  to  the  cube  of  the  second. 

384.  Given  a:b  =  c:  d,  e:f=g:h,  k:l  =  m:n.  Then  aek : 
bfl  =  cgm :  dhn. 

Proof:    Since  §«.£,  5  =  2,    *«»« 

b     d    f    h     I      n 

Multiplying,  J7i=T- 

bfl      dhn 

Or,  aek  :  bfl  =  cgm  :  dhn. 

That  is : 

The  products  of  the  corresponding  terms  of  two  or  more  pro- 
portions are  in  proportion. 

APPLICATIONS  OF  THE  PROPERTIES  OP  PROPORTION 

385.  Of  the  important  properties  of  proportion  in  common 
use  in  the  solution  of  problems  involving  certain  relations,  we 
may  note  briefly  the  following : 

(1)  Since,  in  any  proportion,  the  product  of  the  means  equals  the 
product  of  the  extremes  (Art.  371),  we  may  readily  find  any  one  term 
of  a  proportion  when  three  terms  are  known. 


324  RATIO.     PROPORTION.     VARIATION 

(2)  Since  —  =  -  =  r,  we  have  a  —  br,  and  c=  dr  ;  and  the  substitution 

b     d 
of  these  values  for  a  and  c  will  be  of  frequent  service  in  reductions. 

(3)  Composition  and  division  (Art.  379)  are  of  frequent  use  in  mini- 
mizing the  work  necessary  for  the  solution  of  certain  types  of  equations. 

(4)  An  assumed  identity  involving  any  four  numbers,  a,  6,  c,  and  d, 
is  shown  to  be  true  if,  by  transformations,  we  obtain  a  proportion, 
a:b  =  c:d. 

Illustrations : 

1.  Find  the  ratio  of  x  to  y  when  3a?  +  2y  =  - . 

In  the  form  of  a  proportion  we  have 

Zx  +  2y.4:X-3y  =  5:6. 
By  Art.  371,  6  (3*  4  2  y)  =  5  (4  x  -  3  y) . 

Whence,  18x4l2*/  =  20x-  15  y. 

2x  =  27  y. 
Therefore   (Art.  374),  x  :  y  =  27  :  2.     Result. 

2.  If  a:  6  =  c:  d,  show  that  a  +  3c:  6  +  3<2  =  2a+c  :  26+riL 

Since  a  :  &  =  c  :  d,  we  nave  -  =  -  —  r.     Whence,  a  =  br,  and  c  =  dr. 
6      d 

Reducing  each  ratio  separately, 

a  4  3  c  _  (6r)  +  3  (dr)  =  r  (b  4  3  <T)  = 

643d  6  +  3d         '    6  +  3d5 

2g  +  c=2(&r)+(dr)=r(264d):,r 

2&4d  2&4d         '    26+d 

a  4  3c      2a+c 


Therefore, 


3.    Solve  the  equation 


b+Sd     2&4d 

'     2iC  — 1  X  +  4: 


x2  +  2x  —  l     x2  4- a  4- 4' 
By  composition  and  division  (Art.  379), 

(2  x  -  1)  +  (x2  4  2  s  -  1 )  ■ . .  (x  4  4)  +  (x2  4  x  +  4) 
(2x-l)-(x24  2x-l)      (x  +  4)-(x'2  +  x+4)' 

Simplifying,  x2  +  4x  -  2  =x2  4  2x48^ 

—  x-  —  X2 

Whence,  x244x-2  =  x24  2x4  8. 

2  a;  =  10. 
x  =  5.     Result. 


APPLICATIONS  OF  THE  PROPERTIES  OF  PROPORTION     325 

4.  Find  two  numbers  whose  sum  is  to  their  product  as  3  is 
to  20;  and  whose  sum  increased  by  1  is  to  their  difference 
increased  by  1  as  7  is  to  1. 

Let  x  and  y  =  the  two  required  numbers. 

Then,  x  +  y  :  xy  =  3  :  20.    (By  the  first  condition.)  (1) 

And,  x  +  y  +  l-.x  —  y  +  l=7:l.    (By  the  second  condition.)  (2) 

From  (1),  20  x  +  20  y  =  3xy.  (3) 

From  (2),  3z-4y  =  -3.  (4) 

x  =  *JL=l. 
3 


From  (4), 


Hence,  in  (3),        2o(±l^)  +  20y  =  3y  (±£p^. 

Simplifying,  12  y*  -  149  y  +  60  =  0.  (6) 

Solving  (5),  j      y  =  12,  or  T52. 

x  =  15,  or  f. 

It  will  be  found  that  the  integral  values  only  satisfy  the  given  condi- 
tions, and  the  fractional  values  are,  consequently,  rejected. 
Therefore,  12  and  15  are  the  required  numbers. 


Exercise  122 

Show  that  the  following  are  true  proportions : 

1.  4:8  =  5:10.  4.    3:9.125  =  2:6^. 

2.  2:7  =  2.5:8.75.  5'   **:  3y  =  8a>y :  6^. 

3.  3:5*  =  6:10x.  6'    *>£-*^J? 
Determine  whether  the  following  are  true  proportions  : 

7.  34:53  =  19:39.  9.    12.1 :  4.4  =  2.2  :  .8. 

8.  4£:4  =  8f:8J.  10.    ar>-l :  x  +  1  =  x-l  :  x. 
Find  the  value  of  the  unknown  in  each  of  the  following : 

11.  8:10  =  12:».  13.    32:12  =  x:6. 

12.  4:  z  =  16: 15.  14.   y:  8  =  12:  4. 


18.  4c:a  =  3sc:6a. 

19.  2  am  :  3  =  mx  :  15. 
1 


20. 


a-1 


326  RATIO.    PROPORTION.     VARIATION 

15.    5:  z  =  10.5:. 3. 

13.   2a:5  =  6:3. 

17.   3<c:16  =  .9:.2. 
Find  the  ratio  of  x  to  y  in : 

21.  5x  =  7y. 

22.  3#  + 2?/  =  52/-3a;. 

23.  2  ic  +  y  :  3  x  —  y  =  f. 
Find  the  value  of  x  and  ?/  in : 
27.    a-  l:y  +  1  =  2:3, 


:  a  +  l  =  x  :  a2  —  1 


24. 


2x  +  ?/      m 


3»—  y      ft 

25.  »2-9?/2  =  8iC2/. 

26.  x  —  y:3y  =  y  —  x:x  + 


30.    a  +  1 :  y  + 1  =  3  :  4, 

a :  2  =  y  :  3. 

w    x  +  y+1^2     x+5J 


32. 


a;  4-  V  =  5. 

28.  x  +  2:y  +  2  =  2:3, 
3a-2/  +  l  =  0. 

29.  a;  +  2/ :  «  —  2/  =  3  :  2, 
a;  +  l:2/-l  =  3:2. 

Find  a  mean  proportional  between  : 

33.   20  and  5.  34.    3  and  27.  35.   4  a?x  and  16  aV. 

36.   ^  +  3,  +  2    ^  ,2-,-2 


a  +  2/-l      3'     2/+2 

a;  +  2  y  _  5      a;  + 1  _  44 
a;-22/~6'     y  +  l~2l' 


x2-3x  +  2 
37.    ^  ~  5  +  *  and 


ar*+2a;-3 
ar2 


a;-l 


38. 


2V2 


and 


a;2-l 


5+4V2         4+3V2 
Find  a  third  proportional  to : 

39.  12  and  16.  42.  (c  +  ar)2  and  c2  -  ar*. 

43.  a8  -  1  and  a?  +  a;  +  1. 

40.  3  a2  and  2  a3. 

44.  (m  +  ")2and(m2-n2->. 

41.  2:8  and  3:5.  n4  n2 


APPLICATIONS   OF  THE  PROPERTIES   OF  PROPORTION     327 

Find  a  fourth  proportional  to  : 

45.  4,  7,  and  9.  48.   x2  —  1,  x  + 1,  and  x  —  1. 

46.  3a;,  2  y,  and  62.  49.    c3 -  d3,  c2 -  a*2,  and  c2  -f  cd  +  cP. 

47.  |,  |,  and  |.  50.   1  +  V2,  2  +  V2,  and  2  -  V2. 

Change  the  form  of  each  of  the  following  proportions  so  that 
the  unknown  quantity  shall  occur  in  but  one  term : 

51.  2:5  =  3  — a?:  a;.  56.  m:  7i=p  —  z:z. 

52.  5  :  3  =  4  —  x :  x.  57.  3  :  2  =  x  +  1 :  x. 

53.  6  :  7  =  12  —  y  :  y.  58.  15  :  7  =  x  +  1  :  a;. 

54.  4a  :  3a  =  10  —  a;:  #.  59.  3a  :  5  c  =  x  4- 1  :  x  —  1. 

55.  a3:  a2c  =  c  —  a;:  x.  60.  a2+l :  a2  —  l  =  #-}-l :  &—  1. 

61.  c-\-d:c  —  d  =  c  +  y:c  —  y. 

62.  m24-m  +  l:m2  —  m  —  l  =  a;  —  2:a;  —  8. 

63.  Find  two  numbers  in  the  ratio  of  2  :  3,  such  that  if  each 
is  decreased  by  1  their  ratio  becomes  as  3  :  5. 

64.  Find  two  numbers  in  the  ratio  of  4  :  7,  such  that  if  each 
number  is  increased  by  2  the  ratio  becomes  as  5  :  8. 

65.  If  ad  =  bc,  write  all  the  possible  proportions  whose 
terms  are  a,  b,  c,  and  d. 

66.  Find  two  numbers  whose  sum  is  to  their  product  as 
8  :  15,  and  whose  sum  is  to  their  difference  as  4:1. 

67.  What  number  must  be  added  to  each  of  the  numbers, 
4, 14, 10,  and  30,  that  the  resulting  sums  may  be  in  proportion  ? 

68.  Separate  10  into  two  parts  such  that  their  product  shall 
be  to  the  difference  of  their  squares  as  6  :  5. 

69.  Two  rectangles  have  equal  areas,  and  their  bases  are  to 
each  other  as  5  :  16.     What  is  the  ratio  of  their  altitudes  ? 

70.  The  lengths  of  the  sides  of  three  squares  are  in  the 
ratio  of  2,  3,  and  4.  Find  a  side  of  each  if  the  sum  of  the 
three  areas  is  725  square  units  of  area. 


328  RATIO.    PROPORTION.     VARIATION 

71.  If  a  :  b  =  2  :  3,  and  b  :  c  =  3  :  5,  find  the  ratio  of  a  :  c. 

72.  If  x  :  y  =  3  :  4,  and  y  :  z  =  5  :  6,  find  the  ratio  of  a? :  z. 

73.  If  m  :  n  =  3  :  k,  and  n  :  p  =  k  :  4,   what  is   the   ratio  of 
m  :  p? 

74.  If  c  :  d  =  3.2  :  4.8,  and  d  :  fc  =  3.2  :  8,  find  the  ratio  of 
c  :  k. 

75.  If  a  :  b  =  ft  :  c  =  2  a;  :  3  y,  find  the  ratio  of  a  to  c  in  terms 
of  x  and  ?/. 

76.  If  a  :  ft  =  3  :  4,  find  (3a  +  2&)  :  (2a  +  36). 

77.  If  m  :  w  =  5  :  2,  find  (2  m  +  n)  :  (2  m  —  ?i). 

78.  If  a  :  b  =  2x  :  Sy,  find  (2 a  +  #)  :  (2 a  —  #)  in  terms  of 
b  and  y. 

79.  If  a +  b  :  b  =  x-\-y  :  y,  find  (#  +  2/)  :  (®—y)  in  terms  of 
a;  and  y. 

80.  If  a;  -|-  a  :  #  —  a  =  y  +  x  :  y  —  z,   find  (# -f  ?/)  :  (x  —  y)    in 
terms  of  a  and  z. 

If  a  :  b  —  c  :  d  prove  that : 

81.  ab:cd  =  b2:  d\ 

82.  a  +  c  :  b  +  d  =  c  :  d. 

83.  a2d:b  =  be2  :  d. 

84.  3a2-262:262  =  3ac-26d:  2bd. 


85.  Va2-62:  Vc2-d2  =  6  :  d 

86.  a2  +  2a&  :a&  =  c2  +  2cc?:  cd 
If  a  :  b  =  b  :  c,  prove  that : 

87.  ab-b2:  bc-c?  =  b2:  c2. 

88.  a  —  b  :  6  —  c  =  6  :  c. 

89.  a\a  +  6)  :  c(b  +  c)  =  a2(a  -  6)  :  c(b-  c) . 


90.    a  +  b  :  &  +  c=Va2-&2:  V&  -c2. 


APPLICATIONS   OF   THE  PROPERTIES   OF  PROPORTION     329 

Solve  the  equations : 

91.  2a  +  3  :  2x  +  5=3a+2  :  3a;  +  4. 

92.  2^  +  2^  +  1  :  2a2-2#  +  l:=a  +  l  :  x-1. 

93.  x  +  m  :  x  —  m  =  m-\-n  :  m  —  n. 

94.  x2-l:x  +  l=x2  +  l:x-l. 

95.  x2  +  2x  +  4:  :  x*-x-l  =  x  +  2  :  x-2. 
a?  +  x-l      x?-x  +  2 


96. 
97. 
98. 
99. 
100. 


a^-z  +  a     ^  +  x_2 

a;  +  n  + 1  _  3#— n+1 ' 
x  +  n  —  1      3  a;  —  n  —  1 

s2  +  2a?  +  3==s2  +  3a;-4 

aj2  +  3a;  +  2      z2  +  4a;-2' 

3ss  +  o;  +  l=3s2  — a;— 1 

4:X?  +  X+1       4ar*— a;— 1 


5_Va7+2     6-  Vaj  +  1 


4  +  Va;  +  2     3+Va  +  l 

101.  Separate  100  into  three  parts  which  shall  be  in  the 
Tatio  of  2  :  3  :  5. 

102.  Three  angles  of  a  certain  triangle  are  in  the  ratio  of 
1:2:3.  If  the  sum  of  the  angles  of  a  triangle  is  180°,  find 
the  number  of  degrees  in  each  of  the  angles. 

103.  The  sum  of  three  sides  of  a  triangle  is  240  feet,  and 
the  ratio  of  the  sides  is  as  3:4:  5.  Find  the  length  of  each 
side. 

104.  For  what  value  of  a  will  the  quantity  a  —  1  be  a  mean 
proportional  between  the  quantities  a  +  3  and  a  +  2  ? 

105.  If  2  is  subtracted  from  the  smaller  of  two  numbers 
and  4  is  added  to  the  larger,  the  ratio  is  as  3  :  5,  but  if  1  is 
subtracted  from  the  greater  while  1  is  added  to  the  smaller, 
the  ratio  is  as  10  :  9.     Find  the  numbers. 


330  RATIO.     PROPORTION.     VARIATION 

106.  Two  trains  approach  each  other  between  two  points 
100  miles  apart,  and  their  rates  of  traveling  are  as  4 :  5.  How 
many  miles  will  each  have  traveled  when  they  meet  ? 

107.  Find  two  numbers  whose  sum  is  10,  such  that  the 
ratio  of  the  sum  of  their  squares  to  the  square  of  their  sum  is 
13:25. 

108.  Find  the  length  of  the  two  parts  into  which  a  line  I 
inches  long  is  divided  if  the  parts  are  in  the  ratio  of  a :  b. 

109.  What  must  be  the  value  of  x  in  order  that  2x  —  1, 
2  x  -f 1,  2  x  +  5,  and  2  (2  x  + 1)  may  be  in  proportion. 

110.  Find  each  side  of  a  triangle  whose  perimeter  is  n 
inches,  the  ratio  of  the  sides  being  as  a  :  b :  c.    . 

VARIATION 

386.  A  variable  is  a  quantity  that,  under  the  conditions  of  a 
problem,  may  have  many  different  values. 

387.  A  constant  is  a  quantity  that,  in  the  same  problem,  has 
a  single  fixed  value. 

388.  If  the  ratio  of  any  two  values  of  a  given  variable 
equals  the  ratio  of  any  two  corresponding  values  of  a  second 
variable,  then  the  first  quantity  is  said  to  vary  as  the  second 
quantity. 

Illustration : 

Suppose  an  automobile  is  running  at  a  speed  of  10  miles  per  hour. 
The  total  distance  traveled  at  the  end  of  any  hour  depends  upon  the  total 
number  of  hours  that  have  elapsed  since  the  start  was  made. 

If  it  runs  6  hours,  the  distance  covered  is  60  miles  ; 

If  it  runs  9  hours,  the  distance  covered  is  90  miles. 

Clearly,  therefore,  the  ratio  of  the  two  periods  of  time  is  the  same  as 
the  ratio  of  the  two  distances  covered.    That  is  : 

6  :  9  =  60  :  90. 

We  may  say,  therefore,  that  the  distance  (d)  varies  as  the  time  ($)»  or 

d  cc  t. 

The  symbol  cc  is  read  "varies  as." 


VARIATIONS  UNDER  DIFFERING  CONDITIONS        331 

389.  If  xccy,  then  x  equals  y  multiplied  by  some   constant 
quantity. 

Proof :    Let  a  and  b  denote  any  one  pair  of  corresponding  values  of 
x  and  y. 

Then  by  definition,  -  =  -. 

y     b 

From  which,  %  =  -  y  • 

b 

Denoting  the  constant  ratio,  -»  by  c,    x  =  cy. 
b 

In  general : 

We  may  change  any  variation  to  an  equation  by  the  introduc- 
tion of  the  constant  factor  or  ratio. 

390.  If  one  pair  of  corresponding  values  for  the  variables  in  a 
given  variation  is  known,  the  constant  ratio  is  readily  obtained. 

Illustration : 

If  x  varies  as  y,  and  x  =  12  when  y  =  3,  find  the  value  of  x 
when  y  =  10. 

We  have  x  —  cy. 

Substituting,  12  =  c.  x  8. 

Whence,  c  =  4. 

Hence,  when  y  =  10,  x  =  4  y 

=  4x10 

=  40.    Result. 

VARIATIONS  UNDER  DIFFERING  CONDITIONS 

(a)  Direct  Variation 

391.  The  simple  form,  xccy,  is  a  direct  variation. 

Illustration : 

The  distance  covered  by  a  train  moving  at  a  uniform  rate  of  speed 
varies  directly  as  the  time  elapsed.     That  is  : 

d  x  t  or  d  =  ct,  where  c  is  the  constant  ratio. 


832  RATIO.     PROPORTION.    VARIATION 


(6)  Inverse  Variation 

392.  A  quantity  is  said  to  vary  inversely  as  another  quantity 
when  the  first  varies  as  the  reciprocal  of  the  second.    ■ 

Illustration: 

The  time  (£)  required  by  an  automobile  going  from  New  York  to  Phila- 
delphia varies  inversely  as  the  speed  (s)  ;  for  if  the  speed  is  doubled,  the 
time  required  will  be  but  one  half  the  former  time.     That  is : 

1  c 

tec-  or  t  =  -,  where  c  is  the  constant  ratio. 
s  s 

I 

(c)  Joint  Variation 

393.  If  a  quantity  varies  as  the  product  of  two  quantities, 
the  first  quantity  is  said  to  vary  jointly  as  the  other  two 
quantities. 

Illustration: 

The  number  of  dollars  (N)  paid  to  a  motorman  for  a  certain  number 
of  trips  (£)  varies  jointly  as  the  number  of  dollars  paid  to  him  for  one 
trip  (w)  and  the  number  of  trips  made.    That  is : 

Nccnt  or  ST  as  cnt,  where  c  is  the  constant  ratio. 


(d)  Direct  and  Inverse  Variation 

394.  One  quantity  may  vary  directly  as  a  second  quantity, 
and,  also,  inversely  as  a  third  quantity.  In  such  a  case  the 
quantities  are  said  to  be  in  direct  and  inverse  variation. 

Illustration : 

In  mowing  a  field,  the  time  required  for  the  work  (t)  varies  directly  as 
the  number  of  acres  in  the  field  (A),  but  inversely  as  the  number  of  men 
engaged  at  the  task  (n) .     That  is  : 

A  cA 

t  oc  —  or  t  =  — ,  where  c  is  the  constant  ratio. 
n  n 


VARIATIONS  UNDER   DIFFERING   CONDITIONS        333 


(e)  Compound  Variation 

395.   The  result  obtained  by  taking  the  sum  or  the  differ- 
ence of  two  or  more  variations  is  a  compound  variation. 

Illustration : 

If  y  equals  the  sum  of  two  quantities,  a  and  b,  and  if  a  varies  directly 

as  x8  while  b  varies  inversely  as  x2,  then 

c' 
a  =  ex8  and  b  =  — 

x2 

c' 
Adding  (since  y  =  a  +  6),      y  =  ex8  +  — . 

x'2  * 

It  is  to  be  noted  that  in  such  cases  tw%  different  factors  are  necessary. 


(/)  Important  Principle 

396.  If  x  depends  only  upon  y  and  z,  and  if  x  varies  as  y  ivhen 
z  is  constant,  and  x  varies  as  z  when  y  is  constant,  then  x  varies 
as  yz  when  both  y  and  z  vary. 

Let  x,  y,  z  (1),  Xi,  yi,  z  (2),  and  x2,  jfi,  z\  (3)  be  three  sets  of  corre- 
sponding values.    Then, 

Since  z  has  the  same  value  in  (1)  and  (2), 

Since  yx  has  the  same  value  in  (2)  and  (3), 

Multiplying  (I)  by  (II), 

Or, 
That  is : 

The  ratio  of  any  two  values  of  x  equals  the  ratio  of  the  corre- 
sponding values  of  yz,  and,  by  definition,  x  varies  as  yz. 

Illustration : 

The  area  (A)  of  a  rectangle  varies  as  the  base  (B)  when  the  height  (H) 
is  constant,  and  the  area  varies  as  the  height  when  the  base  is  constant. 
Therefore,  when  both  the  base  and  height  vary,  the  area  will  vary  as  the 
base  and  height  jointly. 


*i    y\ 

(I) 

X\  _  z 
X2        9% 

(II) 

x  _  yz 

x2     y\z\ 

x-.x2  =  yz:  yxz\. 

334  KATIO.    PROPORTION.    VARIATION 

I  II  III  IV 


8 

10 

8 
.  12 

5 
10 

5 
IS 

A  =  BH 
=  105 

=  50 

A  =[BH 

=  12.5 
=  60 

A  =  BH 

=  10-8 
=  80 

A  =  BH 

=  12-8 
=  96. 

In  II.  B  changes,  H constant,  A  changes  (AccB,  H constant). 
In  III.  H  changes,  B  constant,  A  changes  (AxH,  B  constant). 
In  IV.    B  changes,  H  constant,  A  changes    (AccBH,  Band  H  vary). 


APPLICATIONS   OP   VARIATIONS 

397.   Illustrations: 

1.   If  x  varies   as  y2,  and  x  =  2  when  y  —  4,  find  x  when 

y  =  16. 


Since 

xocy2, 

we  have 

a;  =  my2. 

And,  if  x  =  2  and  y  = 

=  4,          2  =  m  x  42. 

Whence, 

m  =  £. 

Then,  when  y  =  16, 

x  =  K16)2. 

»  =  H*. 

x  =  32.     Result. 

2.    If  x  varies  inversely  as  y*,  and  x  =  2  when  y  =  4,  find 

a;  when  y  =2. 

Since 

xoci-, 
y8 

we  have 

Whence, 

2=P' 

and 

m  =  128. 

Then,  when  y  =  2, 

23 

x  =  16.    Result. 

APPLICATIONS  OF  VARIATIONS  335 

3.  If  s  is  the  sum  of  two  quantities,  one  of  which  varies 
directly  as  x2  and  the  other  inversely  as  x,  and  if  s  =  6  when 
x  =  2,  and  s  =  2  when  x  —  —  2,  find  s  when  a;  =  —  1. 

Let  w  and  v  represent  the  two  quantities. 


Then              & 

=  w  +  t>; 

uccx2;            floe-* 
sacs2+-- 

X 

From  which, 

s  =  mx2  +  -• 

(1) 

Substituting : 

Ifs  =  2, 

2                    zJ 

(2) 

If  x  =  -  2, 

2  =  m(-2)2+i  =  4m- 
—  2 

n 
2 

(8) 

From  (2)  and 

(3), 

at  =s  1,  fi  s  4. 

If  x  =  —  1,  in 

(1), 

S  =  l(_l)2+_£_ 

=  1-4 

=  -  3-     Result. 

4.  The  volume  of  a  sphere  varies  as  the  cube  of  its  diame- 
ter. If  three  metal  spheres  whose  diameters  are  6,  8,  and  10 
inches,  respectively,  are  melted  and  recast  into  a  single  sphere, 
what  is  its  diameter  ? 

Let  V  denote  the  volume  of  the  required  sphere,  and  D  its  diameter. 

Then  Foe  2)3. 

Whence,  V=  mD8.  (1) 

Denote  the  volumes  of  the  three  given  spheres  by  Vu  V2,  and  Vs. 

We  have,  therefore,  Vi  =  wi(6)8  =  216  m, 

F2  =  m(8)8  =  512m, 
F3  =  m(10)8  =  1000  m, 
Whence,  by  addition,      V\  +  V2  +  Vs  -  1728  m. 
But,  by  the  conditions,    Vx  +  V2  +  V*  =  V=  m  7)3.        (From  1.) 


336  RATIO.     PROPORTION.    VARIATION 

Hence,  mi)3  =1728  m, 

D3  =  1728, 

D  =12.    Result. 
That  is,  the  diameter  of  the  sphere  obtained  from  the  given  spheres 
is  12  inches. 

Exercise  123 

1.  If  a;  varies  as  ?/,  and  x  =  10  when  y  —  2,  find,  x  when  y  =  5. 

2.  If  a;  varies  as  y,  and  #  =  3.2  when  y  =  0.8,  find  a;  when 
y  =  5.6. 

3.  If  x  +  1  varies  as  y  —  1,  and  cc  =  6  when  y  =  4,  find  03 
when  y  =  7. 

4.  If  2  a;  —  3  varies  as  3  2/  +  2,  and  ?/  =  2  when  x  —  0.2,  find 
y  when  a;  =  1.5. 

5.  If  x2  varies  as  y2,  and  x  =  3  when  y  =  2,  find  y  when 
a;  =  4. 

6.  If  x  varies  inversely  as  y,  and  x  =  2  when  2/  =  4,  find  a; 
when  y  =  S. 

7.  If  a;  varies  inversely  as  y2,  and  x  =  2  when  y  =  ^,  find  y 
when  a;  =  1£. 

8.  If  x  varies  jointly  as  y  and  z,  and  x  =  3  when  y  =  4  and 
2!  =  2,  find  x  when  ?/  =  5  and  2  =  4. 

9.  If  a;  varies  inversely  as  y2  —  1,  and  a;  =  4  when  y  =5, 
find  a;  when  ?/  =  15. 

10.  If  a;  varies  as  -,  and  y  =  —  2  when  a?  =  7,  find  the  equa- 
tion  joining  aj  and  y. 

11.  If  the  square  of  x  varies  as  the  cube  of  y,  and  if  x—  6 
when  2/  =  4,  find  the  value  of  jf  when  x  =  30. 

12.  If  s  is  the  sum  of  two  quantities,  one  of  which  varies 
as  x  while  the  other  varies  inversely  as  x ;  and  if  s  =  2  when 
a?  =  £  and  s  =  2  when  a;  =  —  1,  find  the  equation  between  s  and  x. 


APPLICATIONS  OF  VARIATIONS  337 

13.  If  w  varies  as  the  sum  of  x,  y,  and  z,  and  w  =  4  when 
x  =  2,  y  =  —  2,  and  2  =  5,  find  a;  if  w  =  —  3,  2/  =  2,  and 
2=  -6. 

14.  Given  that  s  =  the  sum  of  three  quantities  that  vary  as 
x,  x2,  and  se3,  respectively.  If  x  =  1,  s  =  3  ;  if  a;  =  2,  s  =  6 ;  and 
if  x  =  4,  s  =  16.     Express  the  value  of  s  in  terms  of  #. 

15.  The  area  of  a  circle  varies  as  the  square  of  its  diameter. 
Find  the  diameter  of  a  circle  whose  area  shall  be  equivalent 
to  the  sum  of  the  areas  of  two  circles  whose  diameters  are  6 
and  8  inches  respectively. 

16.  The  intensity  of  light  varies  inversely  as  the  square  of 
the  distance  from  the  source.  How  far  from  a  lamp  is  a  certain 
point  that  receives  just  half  as  much  light  as  a  point  25  feet 
distant  from  the  lamp  ? 

17.  The  volume  of  a  sphere  varies  as  the  cube  of  its  diameter. 
If  three  spheres  whose  diameters  are  3,  4,  and  5  inches,  respec- 
tively, are  melted  and  recast  into  a  single  sphere,  what  is 
the  diameter  of  the  new  sphere  ? 

18.  The  volume  of  a  rectangular  solid  varies  jointly  as  the 
length,  width,  and  height.  If  a  cube  of  steel  8  inches  on  an 
edge  is  rolled  into  a  bar  whose  width  is  6  inches  and  depth  2 
inches,  what  will  be  the  length  of  the  bar  in  feet  ? 

19.  If  the  amount  earned  while  erecting  a  certain  wall 
varies  jointly  as  the  number  of  men  engaged  and  the  number 
of  days  they  work,  how  many  days  will  it  take  4  men  to  earn 
$  100  when  6  men  working  9  days  earn  $  135  ? 

20.  The  pressure  of  the  wind  on  a  plane  surface  varies 
jointly  as  the  area  of  the  surface  and  the  square  of  the  wind's 
velocity.  The  pressure  on  a  square  foot  is  one  pound  when 
the  wind  is  blowing  at  a  rate  of  15  miles  an  hour.  What 
will  be  the  velocity  of  a  wind  whose  pressure  on  a  square  yard 
is  81  pounds  ? 

SOM.    EL.    ALG. — 22 


CHAPTER   XXV 
•  PROGRESSION 

ARITHMETICAL  PROGRESSION 

398.  A  series  is  a  succession  of  terms  formed  in  accordance 
with  a  fixed  law. 

399.  An  arithmetical  progression  is  a  series  in  which  each 
term,  after  the  first,  is  greater  or  less  than  the  preceding  term 
by  a  constant  quantity.  This  constant  quantity  is  the  common 
difference. 

400.  We  may  regard  each  term  of  an  arithmetical  progres- 
sion as  being  obtained  by  the  addition  of  the  common  difference 
to  the  preceding  term;  hence, 

An  increasing  arithmetical  progression  results  from  a  positive 
common  difference,  and  a  decreasing  arithmetical  progression  re- 
sults from  a  negative  common  difference.     Thus : 

1,  5,  9,  13,  .  .  .  etc.,  is  an  increasing  series  in  which  the  common  differ- 
ence is  4. 

7,  4,  1,  —2,  .  .  .  etc.,  is  a  decreasing  series  in  which  the  common  differ- 
ence is  —3. 

401.  In  general,  if  a  is  the  first  term,  and  d  the  common 

difference, 

a,  a  +  d,  a  +  2  d,  a  4-  3  d,  a  +  4  d,  .  .  .  etc., 

is  the  general  form  of  an  arithmetical  progression. 

338 


ARITHMETICAL  PROGRESSION  339 

402.  The  nth  Term  of  an  Arithmetical  Progression. 

In  the  general  form, 

a,a  +  cf.fl  +  2d,a  +  3d,a  +  4d)  .  .  .  etc., 
it  will  be  seen  that  the  coefficient  of  d  in  any  term  is  less  by  1  than  the 
number  of  the  term.     Hence,  the  coefficient  of  d  in  the  nth  term  is  n  —  1. 
Therefore,  if         a  =  the  first  term  of  an  arithmetical  progression, 
d  =  the  common  difference, 
n  =  the  number  of  terms  in  the  series, 
I  =  the  last  term  (that  is,  any  required  term), 
then,  1  =  a  +  (n  -  l)d.  (I) 

403.  The  Sum  of  the  Terms  of  an  Arithmetical  Progression. 

If  s  denotes  the  sum  of  the  terms  of  an  arithmetical  progression,  we 
may  write  the  following  identities,  the  second  being  the  first  written  in 
reverse  order  : 

s  =  a  +  (a  +  d)  +  (a  +  2 d)  + +  (I  -  2d)  +  (* -  &)  +  I 

8  =  1  +  (l-d)  +  (l-2d)  +  • +  (a  +  2d)  +  (a  +  d)  +  a. 

2  s  =  (a  +  0  +  (a  +  I)  +  (a  +  I)  +  .-•  +  (a  +  1)  +  (a  +  I)  +  (a  +  I). 

Whence,  2s  =  n(a  +  l)> 

Or,  s  =  5(a  +  l).  (II) 

404.  Combining  the  two  formulas,  (I)  and  (II),  we  obtain 
the  following  convenient  formula  for  finding  the  sum  of  an 
arithmetical  progression  when  the  first  term,  the  common 
difference,  and  the  number  of  terms  are  known.     That  is : 

s  =  5[2a  +  (n-l)d].  (Ill) 

z 

405.  The  first  term  (a),  the  common  difference  (d),  the 
number  of  terms  (n),  the  last  term  (I),  and  the  sum  of  the 
terms  (s),  are  the  elements  of  an  arithmetical  progression. 

APPLICATION  OF  THE  FORMULAS  FOR  ARITHMETICAL  PROGRESSION 

406.  Illustrations  : 

1.  Find  the  10th  term  and  the  sum  of  10  terms  of  the 
arithmetical  progression,  1,  4,  7,  10,  .  .  . 


340  PROGRESSIONS 

We  have  given,       a  —  1,  d  =  3,  n  =  10. 
In  the  formula,  I  =  a  +  (n  —  l)d. 

Substituting,  I  =  1  +  (10  -  1)  (3) 

=  28,  the  required  10th  term. 

In  the  formula,  s  =  -  (a  +  Z)- 

2 

Substituting,  s  =  —  (1  +  28) 

z 

=  5  x29 

=  145,  the  required  sum  of  10  terms. 

2.   The  first  term  of  an  arithmetical  progression  is  3,  the  last 
term,  38,  and  the  sum  of  the  terms,  164.     Find  the  series. 

We  have  given,       a  =  3,  I  =  38,  s  =  164. 

Then,  s  =  -  (a  +  Z)    or    164  =  |  (3  +  38),    whence  n  =  8. 

Also,  Z  =  a  +  (w  -  1)  d,   or  38  =  3  +  (8  -  1)  d,     whence  d  =  5. 

Therefore,  3,  8,  13,  18,  23,  28,  33,  38,  is  the  required  series. 

3.    How  many  terms  of  the  series,  2,  5,  8,  •••,  will  be  re- 
quired in  order  to  give  a  sum  of  126  ? 

We  have  a  -  2,  d  =  3,  s  =  126. 

Then    Z  =  a+(n-l)d  =  2  +  (»-l)3  =  2  +  3n-3  =  3n-l. 

This  value  for  Z  in  terms  of  n  is  substituted  in  the  formula 

S=|(a  +  Z). 

Whence,  126  =  2  (2  +  3  n  -  1),  or  3  n2  +  w  =  252. 

2 

Solving  this  quadratic  equation   in  n,  we  have  (using  the  quadratic 
formula), 

n  =  -l±VrT3024-        n  =  -L±JL5,       M  =  9)0r-9|. 
6  6 

Therefore,  the  required  number  of  terms  for  the  given  sum  is  9. 

Formula  III  (§  404)  may  be  used  here  without  finding  Z. 


ARITHMETICAL  PROGRESSION  341 

Exercise  124 

1.  Find  the  16th  term  of  7,  10,  13,  .... 

2.  Find  the  10th  term  of  2,  -  1,  -  4,  .... 

3.  Find  the  12th  term  of  4,  3.2,  2.4,  •••. 

4.  Find  the  10th  term  of  J,  f,  J,  .... 

5.  Find  the  -20th  term  of  f,  f,  |j ..... 

6.  Find  the  10th  term  of  -  12,  -9.5,  -  7,  -4.5,  .... 

7.  Find  the  14th  term  of  x  + 1,  x  +  3,  x  +  5,  •••. 

8.  Find  the  11th  term  of  x  —  5  a,  x  —  4  a,  #—  3  a,  •••. 

9.  Find  the  10th  term  of  4  +  V2,  3  +  2^2,  2  +  3V2,  .... 

10.  Find  the  358th  term  of  .0075,  .01,  .0125,  .015,  .... 
Find  the  sum  of : 

11.  3,  8,  13, 18,  ...  to  24  terms. 

12.  -2,  -9,  -16,  ...  to  12  terms. 

13.  .25,  .3,  .35,  ...  to  40  terms. 
14-   i>lbih  -to  16  terms. 

15.  20V2-10V3, 18V2-9V3, 16V2-8V3,  *.-  to  21  terms. 
Find  the  first  term  and  the  common  difference  when  : 

16.  s  =  297,  w  =  9,  1=16.         19.   Z  =  0,  s  =  100,  n  =  25. 

17.  s  =  294,  7i  =  12,  1  =  41.       20.    n  =  6,  s  =  20 A,  I  =  4.9. 

18.  7i  =  13,  s  =  260,  Z  =  -f.    21.   s  =  0,  Z  =  34.5,  w=24. 
Find  the  common  difference  when: 

22.  a  =  4,  ?  =  40,  n  =  13.  24.   I  =  2,  ti  =  7,  s  =  19.25. 

23.  s  = -27.5,  =  4.5,71=11.   25.    Z  =  .97,  a  =  .8,  s  =  35.4. 

26.  a  =  .08,  s  =  —  25,  7*  =  25. 

27.  7i  =  30,  a=-13V2,  Z  =  16V2. 


342  PROGRESSIONS 

Find  the  first  term  when : 

28.  n  ==  12,  I  =  10,  s  =  60. 

29.  d  =  - 1,  n  =  10,  s  =  -  100. 

30.  d  =  -  .6,  n  =  14,  Z  =  -  .83. 
Find  the  number  of  terms  when: 

31.  a  =  7,  d  =  -3,  Z  =  -23. 

32.  Z  =  f,  a  =  },  s  =  17. 

33.  a  =  -|,Z  =  -A,d  =  -^. 
How  many  consecutive  terms  of 

34.  2,  -  3,  -  8,  •-.  will  give  a  sum  of  -  205  ? 

35.  1,  1J,  If,  •••  will  give  a  sum  of  831  ? 

36.  —  3,  -  3 },  —  4,  •  •  •  will  give  a  sum  of  —  97£  ? 

37.  0.36,  0.32,  0.28,  •••  will  give  a  sum  of  -.4  ? 

38.  12V3, 10V3,  8V3,  will  give  a  sum  of  0? 

THE  DERIVATION  OF  GENERAL  FORMULAS  FROM  THE  FUNDAMENTAL 

FORMULAS 

407.  From  the  fundamental  formulas  (I)  and  (II)  in  Arts 
402  and  403  we  may  derive  a  formula  for  any  desired  element 
in  terms  of  any  other  three  elements. 

Illustrations : 

1.    Given  a,  w,  and  s;  derive  a  formula  for  d. 

From  the  fundamental  formulas, 

l  =  a+(n  —  l)d    and     s=-(a  +  J), 

Ss 

we  must  eliminate  the  element  I ;  and  by  substituting  the  value  of  I  from 
the  first  formula  for  I  in  the  second  formula,  we  have 

s  =-[a  +  a  +  (n-  l)d]. 


ARITHMETICAL  PROGRESSION 


343 


From  which,  2  s  =  ?i[2  a  +  (n  —  l)d], 

2  s  =  2  a?i  +  w(w  —  l)a*, 
2  s  —  2  an  =  n(»  —  l)d", 

d  _  2(g-an)^   the  required  formula  for  d. 

n(n  —  1) 

2.   Given  d,  I,  and  s,  find  a  formula  for  n. 

From  (I),  Art.  402,  I  =  a  +  (n  -  l)rZ. 

Hence,  a  =  Z  —  (ft  —  l)a\ 

Substituting  this  value  for  I  in  (II),  Art  403,  we  have, 

«  =  |[2  a+  (n-  l)d] 

*   =^[2Z-2(n-l)d+(n-l)d]. 

2 

2s  =  2  »Z-n(n-l)d. 
on2-  (2Z  + d>-f  2s  =  0. 


Simplifying, 
Whence, 

Solving  for  n, 


2  Z  +  d  ±  V(2  Z  +  d)2  -  8  ds. 
_ 


1.  Given 

2.  Given 

3.  Given 

4.  Given 

5.  Given 

6.  Given 

7.  Given 

8.  Given 

9.  Given 

10.  Given 

11.  Given 

12.  Given 


Exercise  125 

dj  ft,  and  I ;   derive  a  formula  for  a. 


n,  Z,  and  s 
a,  ft,  and  s 
a,  I,  and  n 
a,  ft,  and  s 
a,  I,  and  s 
dj  I,  and  n 
n,  ?,  and  s 
a,  Z,  and  s 
dj  ft,  and  s 
a,  c?,  and  s 
a,  d,  and  s 


derive  a  formula  for  a. 
derive  a  formula  for  I. 
derive  a  formula  for  d. 
derive  a  formula  for  d. 
derive  a  formula  for  d. 
derive  a  formula  for  s. 
derive  a  formula  for  d. 
derive  a  formula  for  n. 
derive  a  formula  for  a. 
derive  a  formula  for  I. 
derive  a  formula  for  ft. 


344  PROGRESSIONS 


ARITHMETICAL  MEANS 

408.  If  we  know  a  and  b,  the  first  and  last  terms  respec- 
tively in  an  arithmetical  progression,  we  may  form  an  arith- 
metical progression  of  m  +  2  terms  by  inserting  m  arithmetical 
means  between  a  and  b. 

Illustration : 

Insert  7  arithmetic  means  between  3  and  43. 

We  seek  an  arithmetical  progression  of  (m  +  2)  =  (7  +  2)  =  9  terms. 
(For  two  terms,  the  first  and  the  last,  are  given.) 
Therefore,  a  =  3,  I  =  43,  n  =  9.     It  remains  to  find  d. 

In  the  formula,  1=  a+  (n  —  l)d. 

Substituting,  43  =  3  +  (9  —  1)<?. 

Whence,  d  =  5. 

Therefore,  3,  8,  13,  18,  23,  28,  33,  38,  43  is  the  required  series. 

409.  To  insert  a  single  arithmetic  mean,  m,  between  two 
numbers  a  and  b,  we  have  as  a  result  the  series  a,  m,  I. 

Therefore,  m  —  a  =  l  —  m. 

And,  2m  =  a+  I. 

a  +  I 

Or,  the  arithmetic  mean  between  two  numbers  equals  one  half 
the  sum  of  the  numbers. 

Exercise  126 

1.  Insert  5  arithmetical  means  between  13  and  37. 

2.  Insert  9  arithmetical  means  between  6  and  11. 

3.  Insert  7  arithmetical  means  between  f  and  f . 

4.  Insert  15  arithmetical  means  between  —  f  and  f. 

5.  Insert  the  arithmetical  mean  between  12.4  and  13.2. 

6.  Insert  the  arithmetical  mean  between  a  +  2  and  a  —  2. 

7.  Insert  the  arithmetical  mean  between and • 

a+  c         a  —  c 


ARITHMETICAL  PROGRESSION  345 

PROBLEMS  INVOLVING  ARITHMETICAL  PROGRESSIONS 

410.    Illustrations: 

1.  The  4th  term  of  an  arithmetical  progression  is  11,  and 
the  9th  term  26.     Find  the  first  3  terms  of  the  progression. 

The  fourth  term  is  a  +  3  d,  and  the  ninth  term  a  +  8  d. 
Therefore,  a  +  8  d  =  26. 

And,  a  +  3  d  =  11. 

Whence,  5  d  =  15. 

d  =  'S. 

a  =  2. 

Hence,  the  required  terms  of  the  progression  are  2,  5,  8.    Result. 

2.  Find  5  numbers  in  arithmetical  progression,  such  that 
the  product  of  the  1st  and  5th  shall  be  28,  and  the  sum  of 
the  2d,  3d,  and  4th  shall  be  24. 

Let  x  —  2  y,  x  —  y,  x,  x  +  y,  and  x  +  2  y  represent  the  numbers. 

Then,  (x  -  2  y)  (x  +  2  y)  =  28.  (1) 

And,  x-y  +  x  +  x  +  y  =  24.  (2) 


(3) 
(4) 


Hence,  the  required  numbers  are  2,  5,  8,  11,  14  ;  or  14,  11,  8,  5,  2. 
The  symmetrical  forms,  x  —  2  y,  x  —  y,  x,  x  +  y,  x  +  2y,  are  chosen 
merely  for  convenience. 

Exercise  127 

1.  How  many  numbers  between  50   and  500   are  exactly- 
divisible  by  6  ? 

2.  Find  the  sum  of  all  the  numbers  of  two  figures  that  are 
exactly  divisible  by  7. 


From  (2), 

3z  =  24. 

x  =  S. 

From  (1), 

X2_±y2  =  28. 

Substituting  from  (3), 

82  _  4  y2  =  28. 

4  y2  =  36. 

y=±3 

346  PROGRESSIONS 

3.  Find  the  sum  of  the  first  20  odd  numbers. 

4,  Find  x  so  that  2  x  —  1,  2  x  +  2,  3  x  —  2,  and  3  a  + 1  shall 
be  in  arithmetical  progression. 

5    Are  the  3  numbers,  5x  —  3y,  x  +  2y,  and  7y  —  3x,  in 
arithmetical  progression  ? 

6.  What  will  be  the  value  of  x  if  the  numbers  2  x  —  3,  xy 
and  10  —  x  are  in  arithmetical  progression  ? 

7.  Find  the  sum  of  1  +  2  +  3  +  4  +  5+  ..-ton  terms. 

8.  Find  the  sum  of  1+3  +  5  +  7+  •  •  •  to  n  terms. 

9.  Find  the  sum  of  21  terms  of  an  arithmetical  progression 
whose  middle  term  is  23  and  whose  common  difference  is  2. 

10.  Find  the  sum  of  2  +  4  +  6  +  8+  •••to  n  terms,  and 
compare  the  result  with  that  of  example  8. 

11.  Find  the  sum  of  the  first  25  numbers  that  are  divisible 
by  7. 

12.  The  sum  of  3  numbers  in  arithmetical  progression  is 
24,  and  the  product  of  the  2d  and  3d  is  88.  What  are  the 
numbers  ? 

13.  The  5th  term  of  an  arithmetical  progression  of  49 
terms  is  3,  and  the  15th  term,  63.     Find  the  34th  term. 

14.  The  6th  term  of  an  arithmetical  progression  is  —19, 
and  the  sum  of  the  first  18  terms,  36.  Find  the  common  dif- 
ference and  write  the  first  5  terms  of  the  series. 

15.  Prove  that  the  sum  of  n  consecutive  odd  numbers  begin- 
ning with  1  is  n2. 

16.  A  clerk's  salary  is  increased  $  50  every  6  months  for  a 
period  of  8  years.  At  the  end  of  the  3d  year  he  was  receiving 
$1000.  At  what  salary  did  he  begin  and  what  will  he  receive 
during  the  last  half  of  his  8th  year  ? 

17.  Prove  that  the  sum  of  the  terms  of  an  arithmetical  pro- 
gression in  which  a,  n,  and  d  are  all  equal,  is  equal  to —     ~T     - 


ARITHMETICAL  PROGRESSION  347 

18.  Find  4  numbers  in  arithmetical  progression  such  that 
the  sum  of  the  1st  and  3d  shall  be  44,  and  the  product  of 
the  2d  and  4th,  572. 

19.  How  many  strokes  are  sounded  in  24  hours  by  a  clock 
striking  hours  only  ?         '# 

20.  A  boy  saves  25  cents  the  first  week  of  a  new  year,  and 
increases  his  savings  5  cents  each  week  through  the  entire 
year.     How  much  will  he  have  saved  by  December  31st  ? 

21.  The  sum 'of  three  numbers  in  arithmetical  progression  is 
27,  and  their  product,  693.     Find  the  three  numbers. 

22.  Show  that  if  every  alternate  term  of  an  arithmetical 
progression  is  removed,  the  remaining  terms  will  be  in  arith- 
metical progression. 

23.  A  laborer  agreed  to  fill  40  tanks  with  water,  the  tanks 
being  placed  in  a  straight  line  and  at  a  uniform  distance  of 
10  feet  from  each  other.  Each  tank  holds  18  gallons  and  he 
carries  at  each  trip  2  pails  holding  3  gallons  each.  If  the 
source  of  supply  is  10  feet  from  the  first  tank  in  the  row,  how 
far  does  he  travel  before  the  40  are  filled  ? 

24.  A  man  travels  210  miles.  The  first  day  he  goes  12 
miles,  and  he  increases  each  succeeding  day's  distance  by  2 
miles.  How  many  days  will  he  require  to  complete  the 
journey,  and  how  far  will  he  be  obliged  to  go  on  the  last  day  ? 

25.  There  are  2  arithmetical  progressions,  13,  15,  17,  ••• 
and  37,  35,  33,  •••,  in  one  of  which  d  =  2,  and  in  the  other 
d  =  —  2.  What  must  be  the  number  of  terms  for  both  series 
in  order  that  the  sums  for  both  series  shall  be  equal  ? 

26.  A  contractor  failing  to  complete  a  bridge  in  a  certain 
specified  time  is  compelled  to  forfeit  $  100  a  day  for  the  first 
10  days  of  extra  time  required,  and  for  each  additional  day 
beginning  with  the  11th  the  forfeit  is  daily  increased  by  $  10. 
He  loses  in  all  $2550.  By  how  many  days  did  he  overrun 
the  stipulated  time  ? 


348  PROGRESSIONS 


GEOMETRICAL  PROGRESSION 

411.  A  geometrical  progression  is  a  series  in  which  each 
term,  after  the  first,  is  obtained  by  multiplying  the  preceding 
term  by  a  constant  quantity.  ,  This,  constant  number  is  the 
common  ratio. 

Thus,  3,  6,  12,  24,  48,- 

is  a  geometrical  progression  in  which  the  common  ratio  is  2. 

2,  |,  b  i'- 
is  a  geometrical  progression  in  which  the  common  ratio  is  \. 

412.  In  general,  if  a  is  the  first  term  of  a  geometrical  pro- 
gression, and  r  the  common  ratio, 

a,  ar,  ar2,  ar*,  ar*,  ar5,  •••  etc., 
is  the  general  form  of  a  geometrical  progression. 

413.  The  /7th  Term  of  a  Geometrical  Progression. 

In  the  general  form, 

a,  ar,  ar2,  arz,  ar*,  ar5,  •••  etc., 
it  will  be  seen  that  the  exponent  of  r  in  any  term  is  less  by  1  than  the 
number  of  the  term.     Hence,  the  exponent  of  r  in  the  nth  term  is  n  —  1. 
Therefore,  if  a  =  the  first  term  of  a  geometrical  progression, 
r  =  the  common  ratio, 
n  =  the  number  of  terms  in  the  series, 
I  =  the  last  term  (that  is,  any  required  term). 
Then,  l  =  ar»-1.  (I) 

414.  The  Sum  of  the  Terms  of  a  Geometrical  Progression. 

If  s  denotes  the  sum  of  the  terms  of  a  geometrical  progression,  we 
may  write 

s  =  a  +  ar-\-  ar2  +  •••  arn~s  +  arn~2  +  ar11'1.  (1) 

Multiplying  (1)  by  r, 

rs  =  ar  +  ar2  +  ar*-\ —  arn~2  -f  ar"-1  +  ar".  (2) 

Subtracting  (1)  from  (2), 

rs  —  s  =  ar*1  —  a. 

From  which,  s  =  arW~a.  (II) 


GEOMETRICAL  PROGRESSION  349 

415.  Combining  the  two  formulas  (I)  and  (II),  we  obtain 
the  following  convenient  formula  for  finding  the  sum  of  a 
geometrical  progression  when  the  first  term,  the  last  term,  and 
the  common  ratio  are  known.     That  is : 

s=7fr  cm) 

416.  The  first  term  (a),  the  common  ratio  (r),  the  number 
of  terms  (n),  the  last  term  (I),  and  the  sum  of  the  terms  (s), 
are  the  elements  of  a  geometrical  progression. 

APPLICATIONS  OF  THE  FORMULAS  FOR  GEOMETRICAL  PROGRESSION 

Illustrations : 

1.  Find  the  8th  term  and  the  sum  of  8  terms  of  the  pro- 
gression, 2,  6,  18,  54,  ...  . 

We  have  given,  a  =  2,  r  =  3,  n  =  8. 

In  the  formula,  I  =  arn~\ 

we  substitute  I  =  (2)(3)7. 

Whence,  =  2  •  2187, 

Or,  =  4374,  the  required  8th  term. 

rl  —  a 


In  the  formula, 


r-l 


we  substitute  g  =  (3)  (4374)  -  2# 

3-1 
Whence,  s  =  6560,  the  required  sum  of  8  terms. 

2.   Find  the  12th  term  and  the  sum  of  12  terms  of  6,  —  3, 

We  have  given,  a  =  6,  r  =  —  \,  n  =  12. 

In  the  formula,  I  =  at— \        \  =  6(-  |)H  =  6(-^)  =  -  T&?. 

In  the  formula,  s  =  ^-=1^. 
r-l 

-i-l  -  |  1024 "    t7f¥" 

Hence,  the  required  12th  term  is  -  y^j,  and  the  sum  of  12  terms, 

3mt- 


350  PROGRESSIONS 


Exercise  128 


1.  Find  the  7th  term  of  2,  6,   18,  ...  . 

2.  Find  the  8th  term  of  3,  -  6,  12,  ...  . 

3.  Find  the  6th  term  of  -  3,  -  6,  - 12,  ...  . 

4.  Find  the  9th  term  of  J,  £j  1,  ...  . 

5.  Find  the  9th  term  of  —  f,  f ,  —  £,  ...  . 

6.  Find  the  8th  term  of  3,  3^2,  6,  ...  . 

7.  Find  the  6th  term  of  2.4,  0.24,  0.024,  ...  . 

8.  Find  the  10th  term  of  0.0001,  0.001,  0.01, 

9.  Find  the  15th  term  of  a4x,  aV,  a2x\  ...  . 
Find  the  sum  of : 

10.  2,  4,  8,  ...  to  8  terms. 

11.  40,  20,  10,  ...  to  8  terms. 

12.  120,12,1.2,  ...to  6  terms. 

13.  -§,  I,  -§,  ...  to  5  terms. 

14.  -  27,  9,  -  3,  ...  to  6  terms. 

15.  2J,  1£,  fy  ...  to  10  terms. 

16.  1  +  m  +  m2  +  m3  +  ...  to  n  terms. 

17.  64-32  +  16-8  +  4+  ...  ton  terms.. 
Find  the  common  ratio  when : 

18.  a  =  4,  Z  =  1024,  n  =  9. 

19.  I  =  -  729,  a  =  3,  n  =  6. 

20.  a  =  -  4,  2  =  -  512,  s  =  -  1020. 

21.  a  =  4,  Z  =  -Ti¥,  *  =  ffi. 

22.  a  =  4,  l=s\,  n  =  9. 

23.  0  =  3,*  =  ^,*  =  ^. 

24.  n  =  10,l  =  ^,a  =  2. 

25.  ,9  =  -V^a  =  3,*=--Tk- 


GEOMETRICAL  PROGRESSION  351 

Find  the  number  of  terms  when: 

26.  a=-S,  r=-2,  $  =  1023. 

27.  r  =  —  2,  s  =  —  22,  a  =  —  2. 

28.  a=-2,  Z=-128,  r  =  2. 

29.  Z=-160,  a  =  5,  r=-2. 

30.  s=,m^,r  =  i,l  =  ^. 

31.  a  =  .2,  s  =  204.6,  Z  =  102.4. 

How  many  terms  of  the  series : 

32.  f,  \y  y1^,  •••  will  make  a  sum  of  ffj-? 

33.  18,  —  6,  2,  •••  will  make  a  sum  of  4^4-  ? 

34.  V2,  2,  2 V2,  .-  will  make  a  sum  of  62  +  31 V2  ? 

GENERAL  FORMULAS  DERIVED  FROM  THE  FUNDAMENTAL  FORMULAS 

417.  From  the  fundamental  formulas  (I)  and  (II)  in  Arts. 
413  and  414  we  may  derive  a  formula  for  any  desired  element 
in  terms  of  any  other  three  elements. 


Illustrations : 

1.   Given  a,  I, 

and  s; 

derive 

a  formula  for 

r. 

From  (III)  (Art.  415), 

r  —  1 

we  obtain 

rs  - 

-8  =  rl  —  a. 

Whence, 

rs  - 
r(s  - 

rl  =  s  —  a, 
Z)  =  s  —  a, 

s  -I 

2.   Given  a,  n, 

and  Z; 

derive 

a  formula  for 

s 

From  Ex.  1, 

s-l 

From  (I)  (Art.  4 

US),  l  = 

I         n-1  / 
arnl,  r"-1  =-,  r  —    a/ 

I 

(1) 

(2) 


352 

From  (1)  and  (2), 


PROGRESSIONS 


a_n~i/l 


■\Ta 


C~t/a 


s-  I 

*V+  =  sn~irl  -n-Vl», 
n-y/i)  =  n-l/a»-n~\/T», 


n~l/a 


-in 


Note.     The  general  formulas  for  n  involve  logarithms,  and  are  ordi- 
narily reserved  for  advanced  students. 


Exercise  129 


1.  Given  r, 

2.  Given  r, 

3.  Given  a, 

4.  Given  a, 

5.  Given  r, 

6.  Given  r, 

7.  Given  r, 

8.  Given  a, 

9.  Given  n, 
10.  Given  n, 


n,  and  I 
n,  and  s 
n,  and  I 
r,  and  s 
n,  and  s 
I,  and  s 
n,  and  I 
n,  and  s 
I,  and  s 
I,  and  s 


derive  a 
derive  a 
derive  a 
derive  a 
derive  a 
derive  a 
derive  a 
derive  a 
derive  a 
derive  a 


formula  for  a. 
formula  for  I. 
formula  for  r. 
formula  for.  I. 
formula  for  a. 
formula  for  a. 
formula  for  s. 
formula  for  r. 
formula  for  a. 
formula  for  r. 


THE  INFINITE  DECREASING  GEOMETRIC  SERIES 

418.  If  the  absolute  value  of  r  in  a  geometrical  progression, 
a,  ar,  ar2,  •••,  ar71"1,  is  less  than  1,  the  successive  terms  become 
numerically  less  and  less;  hence,  by  taking  a  sufficiently  great 
number  of  terms,  that  is,  by  making  n  sufficiently  great,  the  nth 
term  becomes  as  small  as  we  may  choose,  although  never  equal 
»to  zero. 


GEOMETRICAL   PROGRESSION  353 

Thus,  if  a  =  1  and  r  =  £,  the  series,  £,  J,  |,  1"5,  •••,  may  be  continued 
until  the  nth  term  is  so  small  as  to  have  no  assignable  value.  Hence,  we 
may  say  that  0  is  the  limiting  value  of  the  nth  term. 

It  follows,  therefore,  that  if  I  approaches  0,  rl  approaches  0.     Hence, 

$  =  rl  ~  a,  or  a  ~      approaches  -^—  as  a  limiting  value. 
r  —  1         1  —  r  1  —  r 

That  is,  s  =  — ^—  is  the  formula  for  the  sum  of  the  terms  of  an  infinite 
1-r 
decreasing  geometric  series.  v 

Illustration : 

Find  the  sum  of  the  infinite  series,  l+i  +  -g-  +  -^H • 

We  have  a  s=  I,  r  =  |. 

Hence,  s  =  -i—  =  -  =  -  •      Result. 

THE  RECURRING  DECIMAL 

419.  A  recurring  decimal  is  the  sum  of  an  infinite  decreasing 
geometrical  progression  in  which  the  ratio  is  0.1  or  a  power  of 
0.1. 

Thus,  .272727  ...  =  .27  +  .0027  +  .000027  +  •••  is  a  geometrical  pro- 
gression in  which  a  =  .27  and  r  =  .01. 

Illustration : 

Find  the  value  of  .292929  — . 

We  have         .292929  •••  =  .29  +  .0029  +  .000029  +  .... 
From  which,  a  =  .29  and  r  =  .01. 

Hence,  s  =  _«_  = -J*?— =  i2^  -  2?.      Result- 

l_r      1-.01      .99     99 

In  case  the  recurring  portion  of  the  decimal  does  not  include 
the  first  decimal  figures,  the  sum  of  the  recurring  portion  is 
found  as  above  and  then  added  to  the  other.     Thus : 

2.23189189  •••  =  2.23  +  .00189189  ... . 

For  the  sum  of  .00189  +  .00000189  +  ■  •-, 

™w*  «  .00189       .00189       189  7 

we  nave  s  = = = = = . 

l_r      1-.001        .999        99900      3700 

Hence,  2.23189189-..  =2  +  ^  + 77fo  =  2#&.     Result. 

SOM.    EL.    ALG.  23 


354  PROGRESSIONS 

GEOMETRICAL  MEANS 

420.  If  we  know  a  and  I,  the  first  and  last  terms  respectively 
in  a  geometrical  progression,  we  may  form  a  geometrical  pro- 
gression of  m-f-2  terms  by  inserting  m  geometrical  means 
between  a  and  I. 

Illustration : 

1.   Insert  4  geometrical  means  between  2  and  -g-J^. 

We  seek  a  geometrical  progression  of  (m  +  2)  =  (4  +  2)  =  6  terms. 

Hence,  a  =  2,  Z  =  ^f  3,  n  =  6. 

Then,  I  =  Of*-*,        ^  =  2.  r6-i,        -2l¥  =  r6,        *  =  r. 

Hence,  the  required  progression  is  2,  f,  f,  A,  ^2r,  Y|f. 

421.  To  insert  a  single  geometrical  mean,  m,  between  the  two 
numbers,  a  and  I,  we  require  the  value  of  m  in 

a  :m  =  m  :l. 
From  which,  m  =  Val. 

The  geometrical  mean  between  two  numbers  equals  the  square 
root  of  their  product. 

PROBLEMS  INVOLVING  GEOMETRICAL  PROGRESSION 

422.  Illustrations: 

1.  The  3d  term  of  a  geometrical  progression  is  2,  and  the 
7th  term  162.     Find  the  5th  term. 

We  have  ar2  -  2,  (1) 

ar6  =  162.  (2) 

Dividing  (2)  by  (1),  r4  =  81,        r  =  3. 

Therefore,        ar2  =  2,  a(3)2  =  2,  a  =  f . 
Hence,  the  5th  term,  ar4,  =  (f )  (3)*  =  (2)  (&)  =  18.    Result. 

2.  Find  3  numbers  in  geometrical  progression,  such  that  the 
sum  of  the  1st  and  3d  decreased  by  the  2d  shall  be  7,  and  the 
sum  of  the  squares  of  the  3  shall  be  91. 


GEOMETRICAL  PROGRESSION  355 

Let  a,  ar,  and  ar2  represent  the  3  numbers. 

Then,  a  —  ar  +  ar2  =  7,  (1) 

a2  +  a*r*  +  a¥  =  91.  (2) 

Dividing  (2)  by  (1),  a  +  ar  +  ar2  =  13.  (3) 

Subtracting  (1)  from  (3),  2  ar  =  6. 

Whence,  ar  =  3, 

3 
r  =  -. 

a 

Substituting  in  (1),  a  -  alA  +  «f- J   =7, 

a_3  +  ?_=7. 
a 

a  =  1  or  9. 

Hence,  the  required  numbers  are  1,  3,  and  9.     Result. 

i 

Exercise  130 

1.  Find  the   first  3   terms   of  a  geometrical   progression 
whose  3d  term  is  9  and  whose  6th  term  is  243. 

2.  Find   the   first   2   terms   of  a  geometrical  progression 
whose  5th  term  is  £  and  whose  10th  term  is  16. 

3.  Insert  4  geometrical  means  between  1  and  243. 

4.  Determine  the  nature,  whether  arithmetical  or  geometri- 
cal, of  the  series,  \,  i,  -J,  ...  . 

5.  Find  the  first  2  terms  of  a  geometrical  progression  in 
which  the  5th  term  is  £  and  the  12th  term  16. 

6.  Which  term  of  the   geometrical  progression,  3,  6,  12, 
...,  is  3072  ? 

7.  Find  to  n  terms  the  sum  of  the  series,  1,  3,  9,  27,  ...  . 

8.  Insert  4  geometrical  means  between  -^  and  32. 

9.  Find,  to  infinity,  the  sum  of  2,  1|,  |,  ...  . 

10.  Find  the  value  of  the  recurring  decimal,  0.1515  ...  . 

11.  Find  the  value  of  x  such  that  x  —  1,  x  +  3,  x  -f- 11  may- 
be in  geometrical  progression. 


356  PROGRESSIONS 

12.  Insert  a  single  geometrical  mean  between  6f  and  5\. 

13.  Find  the  value  of  the  recurring  decimal,  2.214214  ...  . 

14.  Find  4  numbers  in  geometrical  progression  such  that 
the  sum  of  the  1st  and  3d  shall  be  15,  and  the  sum  of  the  2d 
and  4th,  30. 

15.  Find  to  infinity  the  sum  of  4,  —  f,  £,  ...  . 

16.  Find  to  infinity  the  sum  of  1.2161616  ...  . 

17.  Insert  a  single  geometrical  mean  between  3  V2  —  2  and 

3V2  +  2. 

18.  The  1st  term  of  a  geometrical  progression  is  10,  and 
the  sum  of  the  terms  to  infinity  is  20.  Find  the  common 
ratio. 

2       11 

19.  Find  to  infinity  the  sum  of  — — >  — =>  — —•>  •••  . 

J  V2  V2  2V2 

20.  Insert  3  geometrical  means  between  a~*  and  a4,  and 
find  the  sum  of  the  resulting  series. 

o  x 

21.  Insert  5  geometrical  means   between  —3  and  —  and  find 

the  sum  of  the  series.   * 

22.  What  must  be  added  to  each  of  the  numbers,  5,  11,  23, 
that  the  resulting  numbers  may  be  in  geometrical  progression  ? 

23.  The  sum  of  the  first  3  terms  of  a  decreasing  geomet- 
rical progression  is  to  the  sum  to  infinity  as  7 :  8.  Find  the 
common  ratio. 

24.  If  the  numbers,  x  —  2,  2x  —  l,  and  5  x  +  2,  are  in  geo- 
metrical progression,  what  is  the  common  ratio  of  the  series  ? 

25.  The  sum  of  the  first  8  terms  of  a  geometrical  pro- 
gression is  equal  to  17  times  the  sum  of  the  first  4  terms. 
Find  the  common  ratio. 

26.  The  population  of  a  certain  city  is  312,500,  and  it  has 
increased  uniformly  by  25  %  every  3  years  for  a  period  of 
12  years.     What  was  the  population  12  years  ago  ? 


GEOMETRICAL  PROGRESSION  357 

27.  A  ball  on  falling  to  the  pavement  rebounds  ^  of  the 
height  from  which  it  was  dropped,  and  it  continues  to  succes- 
sively rebound  ^  of  each  preceding  distance  until  it  is  at  rest. 
If  the  height  from  which  it  originally  fell  was  60  feet,  through 
how  great  a  distance  does  it  pass  in  falling  and  rebounding  ? 

28.  What  is  the  condition  necessary  that  a  + 1,  a-j-S,  a+  7, 
and  a  +  15  shall  be  in  geometrical  progression?  For  what 
value  of  a  is  this  condition  true  ? 

29.  Show  that  if  4  numbers,  m,  n,  x,  and  y,  are  in  geo- 
metrical progression,  then  m  +  n,  n  +  a?,  and  x  +  y  are  also  in 
geometrical  progression. 

30.  A  sum  of  money  invested  at  6  %  compound  interest  will 
double  itself  in  12  years.  What  will  be  the  amount  of  $10 
invested  at  compound  interest  at  the  end  of  60  years  ? 

31.  If  4  numbers  are  in  geometrical  progression,  the  sum 
of  the  2d  and  4th  divided  by  the  sum  of  the  1st  and  3d  is 
equal  to  the  common  ratio. 

32.  Find  3  numbers  in  geometrical  progression  whose  sum  is 
21,  and  the  sum  of  whose  squares  is  189. 

33.  Find  an  arithmetical  progression  whose  first  term  is  2, 
and  whose  1st,  3d,  and  7th  terms  are  in  geometrical  pro- 
gression. 

34.  If  the  alternate  terms  of  a  geometrical  progression  are 
removed,  the  remaining  terms  are  in  geometrical  progression. 

35.  Find  to  infinity  the  sum  of  the  series 


X  +  1        \X  +  1J         \x  +  1 

36.  Find  the  4th  term  of  an  infinite  decreasing  geometric 
series  the  sum  of  whose  terms  is  ff>  and  whose  first  term 
is  .25. 


CHAPTER  XXVI 

THE  BINOMIAL  THEOREM.    POSITIVE  INTEGRAL 
EXPONENT 

423.  A  finite  series  is  a  series  having  a  limited  number  of 
terms. 

424.  The  binomial  theorem  is  a  formula  by  means  of  which 
any  power  of  a  binomial  may  be  expanded  into  a  series. 

425.  By  actual  multiplication  we  may  obtain : 

(a  +  by  =  a2  +  2  ab  +  62. 

(a  +  6)3  =  a3  +  3  a26  +  3  a&2  +  bs. 

(a  +  by  =  a*  +  4  a3&  +  6  a262  +  4  a&3  +  6* ;  etc. 

In  the  products  we  observe : 

1.  The  number  of  terms  exceeds  by  1  the  exponent  of  the 
binomial. 

2.  The  exponent  of  a  in  the  first  term  is  the  same  as  the 
exponent  of  the  binomial,  and  decreases  by  1  in  each  succeed- 
ing term. 

3.  The  exponent  of  b  in  the  second  term  is  1,  and  increases 
by  1  in  each  succeeding  term  until  it  is  the  same  as  the  expo- 
nent of  the  binomial. 

4.  The  coefficient  of  the  first  term  is  1,  the  coefficient  of  the 
second  term  is  the  same  as  the  exponent  of  the  binomial. 

5.  If  the  coefficient  of  any  term  is  multiplied  by  the  exponent 
of  a  in  that  term,  and  the  product  is  divided  by  the  exponent 
of  b  in  that  term  increased  by  1,  the  result  is  the  coefficient  of 
the  next  following  term. 

358 


THE   BINOMIAL  THEOREM  359 

426.   By  observing  these  laws  we  may  write  the  expansion 

of  (a  +  6)4thus: 

/     ,  ni       i(.-ijn'*'*Ji'j  4.3-2^,3  .  4.3.2.1M 
(a  +  6)4  =  a4  +  4  «36  +  __a262  +  ^^a&B  +  j^-M 

In  like  manner,  we  may  write  the  expansion  of  (a  +  b)n  in 
the  form : 

(a  +  &)•  =  a»  +  na-ift  +  n("  -  1)  an-262  +  n(n-l)(n-  2)g^ft,  +  ... 
1 «2  1 • 2«  o 

This  expression  is  the  binomial  formula,  and  we  will  now 
prove  that  it  is  a  general  expression  for  any  power  of  (a  -+-  6), 
for  positive  integral  values  of  n. 


PROOF  OP  THE  BINOMIAL  THEOREM  FOR  POSITIVE  INTEGRAL  EXPONENTS 

427.  We  have  shown  by  actual  multiplication  that  the  laws 
governing  the  successive  expansions  of  (a-f  b)  are  true  up  to 
and  including  the  fourth  power. 

If,  now,  we  assume  that  the  laws  of  Art.  425  are  true  for 
any  power,  as  the  nth  power,  and  if,  furthermore,  we  show  the 
laws  to  hold  for  the  (n  -f-  l)th  power,  then  the  truth  of  Art.  425 
for  all  positive  integral  values  of  n  is  established.  This  method 
of  proof  is  known  as  mathematical  induction. 

Both  members  of  the  formula  (1)  below  are  multiplied  by  (a  +  5). 

(a  +  by  =  a»  +  nfl»-*6+  n(n  "  ^«-%*  4-  n<n  ~  &g  ~2)  a»-'6»+  -  (1) 
1  •  2  1  *  2  •  o 

(a +  6)       (a +  6) 

(a  +  &)•+!  =  *h-i  +  na»6  +  "("-^o-iy  +  ^-1)(^-2)aa-2&3  +  ... 

1  •  2  1  « 2  •  o 

+   a*6  +  wa"-1^2  +  w(n-l)  a«-263  +  - 

1*2 

(a  +  6)«+i  =  tfH-i  +  (n  +  l)a»6  +  f"71^-1)  +  »1  a»-i&» 

rn(n-l)(n-2)      n(n -1)"1  w_2&8  ,  ... 
^L         1-2-3  1-2     J 


360  THE   BINOMIAL  THEOREM 

m  tfH-1  +  (n  +  l)a-6  +  QL±^a»-W  +  (»  +  1M;-  1)^268.  +  ....    (2) 
1  •  2  1  •  2  *  O 

It  will  be  observed  that  (2)  is  identical  with  (1)  excepting 
that  every  n  of  (1)  is  replaced  by  n  -j- 1  in  (2).  That  is,  we 
assumed  the  laws  of  425  to  be  true  for  n,  and  have  shown  them 
to  be  true  for  n  +  1.  Similarly,  we  might  show  the  laws  to  be 
true  for  n  -f-  2,  and  so  on,  indefinitely.  Hence,  the  laws  of  Art. 
425  being  true  for  the  4th  power  may  be  shown  to  hold  true  for 
the  5th  power ;  and  holding  for  the  5th  power,  may  be  shown 
to  hold  for  the  6th  power. 

Therefore,  for  any  positive  integral  values  of  n : 

(a  +  b)»  =  a«  +  na*-*b  +  HCEzJQ  a«-2b2  +  n(n-l)(n-2)  ari_31)8 
1-2  1-2-3 

428.  The  Factorial  Denominator. 

In  practice  it  is  convenient  to  write  (_3_  f  or  1  •  2  •  3,  [6_  for 
1 . 2  •  3  •  4  •  5  •  6,  etc. 

We  read  [3  as  "factorial  3,"  [6_  as  "factorial  6,"  etc.  In 
general,  \n_  means  the  product  of  the  natural  numbers 
1  .  2  •  3-4-5  •  •  •  n  inclusive. 

429.  An  expansion  of  a  binomial  is  a  finite  series  when  n  is 
a  positive  integer.  For,  in  the  coefficients,  we  finally  reach  a 
factor,  n  —  n,  or  0.  And  the  term  containing  this  0  factor 
disappears.  Moreover,  every  following  term  contains  this  0 
factor,  hence  each  term  following  disappears. 

430.  The  Signs  of  the  Terms  in  an  Expansion. 

If  both  a  and  b  are  positive  in  (a  +  6)n,  the  signs  of  all  the 
terms  in  the  expansion  are  positive. 

If  b  is  negative,  that  is,  given  (a—  b),  all  terms  involving 
even  powers  of  —  b  are  positive,  while  all  terms  involving 
odd  powers  of  —  b  are  negative.  Therefore,  the  signs  of  the 
terms  of  the  expansion  of  (a— b)n  are  alternately  +  and  — . 


APPLICATIONS   OF  THE   BINOMIAL   FORMULA         361 
APPLICATIONS  OF  THE  BINOMIAL  FORMULA 

431.    Illustrations : 
1.   Expand  (2a  +  3  6)5. 

(2  a  +  3  by  =  (2  ay +5(2  a)*  (8  &)  +  ^J  (2  a)«(8  6)2+  f^|  (2  a)«(S  6)3 

1  •  J  1  •  1  •  o 

1.2.8*4N      yv  1.2.3.4.5^     ' 

=  32  a5  +  240  a46  +  720  a862  +  1080  a*&  4-  810  a6*  +  243  b5.     Result. 


2.    Expand  /^-^Y. 


Changing  to  a  form  best  suited  to  the  binomial  formula, 

(2  er*-af)«=(2  a-!)6-6(2  a-*)*(rft)  +  15(2  a-1)*(a*)2-20(2  a-1)8^)* 

+  15  (2  a-i)2(at)4  -  6  (2  a-1)  (a*)5  +  (a§)« 

=64  a"6  -  192  a-5at  +  240  a-*a%  -  160  a~3a2  +  60  a~H%  -  12  a-^r+a* 
=  64_192  +  240_160  +  60a|_12a|  +  a4     ^ 
a<J     aV      a§       a 

432.    To  find  Any  Required  Term  of  (a  +  b)n. 

In  finding  the  rth  or  general  term  in  an  expansion,  we  ob- 
serve the  general  formation  of  any  term  of  (a  -4-  b)n. 

If  r  be  the  number  of  the  term  required : 

1.  The  exponent  of  b  is  less  by  1  than  the  number  of  the 
term. 

2.  The  exponent  of  a  is  n  minus  the  exponent  of  b. 

3.  The  last  factor  in  the  numerator  of  the   coefficient  is 
greater  by  1  than  the  exponent  of  a. 

4.  The  last  factor  in  the  denominator  of  the  coefficient  is 
the  same  as  the  exponent  of  b. 


362  THE  BINOMIAL  THEOREM 

Therefore,  In  the  rth  term : 

The  exponent  of  b  =  r  —  1. 

The  exponent  of  a  =  n  —  (r  —  1)  =  n  —  r  +  1. 

The  last  factor  of  the  numerator  in  the  coefficient  =  n  —  r  +  2. 

The  last  factor  of  the  denominator  in.  the  coefficient  =  r  —  1. 

Or,  the  rth  term  =  n(n  -  l)(n-2)...  (n -r  +  S) ^-^^ 

IL=1 
Illustration : 

1.  Find  the  7th  term  of  (2  a2  -  a;"1)10. 

We  have  r  =  7,  n  =  10. 

Then,  (2  x2  -  ar1) 10  =  [ (2  x2)  +  (  -  ar*) ]10, 

whence,  for  the  formula  of  Art.  360, 

a  =  (2 ce2)  and  6  =  (-  ar1). 
The  exponent  of6  =  r-l  =  7-l=6. 

The  exponent  ofa  =  w  —  r+l  =  10  —  7  +  1=4. 
The  last  factor  of  the  numerator  of  the  coefficient 

=  w~r-f2  =  10-7+2  =  5. 
The  last  factor  of  the  denominator  of  the  coefficient 

=  r_l=7_l=6. 

Then  the  7th  term      =  10  •  9  ■  8  .  7  ■  6  •  5  (2  x2)4(_  x-i)6 
1.2.3.4.5.6  v :  •■   '  v  } 

=  210  •  16  x8  •  x-* 

=  3810  x2.     Result. 

The  same  principle  enables  us  to  find  the  number  of  a  term  containing 
a  required  power  of  a  letter  when  that  letter  occurs  in  both  terms  of  the 
given  binomial. 

2.  Find  the  term  containing  a11  in  [2  sc8 ]    . 

Let  r  =  the  number  of  the  required  term.  Disregarding  all  but  the 
literal  factors  of  the  term,  we  may  write : 

X11  =  (z3)12-'-+1(ar-2)r-1  =  (a;86-8r+8)  (a;-2r+2)  _  ^l-Sr.. 

Therefore,  r  s=  6,  the  number  of  the  required  term. 

That  is,  on  writing  the  6th  term,  we  shall  find  the  exponent  of  x  to  be  11. 


APPLICATIONS  OF  THE   BINOMIAL  FORMULA        S6S 


Exercise  131 

Expand : 

1.    (a  +  x)\  16.   (ax-1  —  VoaF1)5. 

2-    0-5)4.  17.   (2af*-3aTt)<\ 

3.  (1-2  a)5-  g  y_ 

4.  (3V-2y)»  18'   (2^-W-l)5. 

5.  (4a-<.  ^         4 

6.  (2m2-3^.  19-    (^— ~5j' 

7.  (5**+3*)»  /3Va;__2a^ 

8.  (2a^-32/)7.  ^    ^    a        v|, 

9.  (a+Va)6.  /aV=l  ,        3 


22. 


10.  («Va5  4-^)6. 

11.  (3V^l-2)5. 

12.  (4V2-3V3)«. 

13.  (2aVx-Vx2)*.  23.    U-*yJ?-x-W-a) 


/oV-1  3      \§ 

\2Vic       y 


-  Nl-2^)4 


14.  (3  a  -2-V-  a2)4. 

15.  (2mV2n-3)5. 

Find  the 

25.   5th  term  of  (a  +  a)7.  26.   8th  term  of  (aV-  ax)1*. 


27.  7th  term  of  ( V3a  +  » V2a)10. 

28.  6th  term  of  (f  V*  -  x^V^l)9. 
Find  the  number  of  the  term  containing 

29.  a8  in  (x>  +  x-1)10.  32.   Xs  in  (^  +  3  aj"Y- 

30.  a12  in  (2  *"  -jy    '  33.    ar16  in  A*  or2  -  |Y°  • 

31.  xmfex-  ?Y3  •  34.   x~»  in  f  3  a;"1  -  ~r  J  • 


CHAPTER    XXVIT 
LOGARITHMS 

433.  By  means  of  the  exponents,  2,  3,  4,  etc.,  we  may  express 
certain  numbers  as  exact  powers  of  10. 

Thus,  100  =  102, 

1000  =  103, 
10000  =  10*,  etc. 

434.  Suppose,  now,  that  10  is  given  any  real  exponent,  as  x. 
Then  some  positive  number,  N}  results  as  the  #th  power  of  10. 

That  is,  N=ltf. 

435.  If,  therefore,  we  knew  the  necessary  approximate  values 
for  x,  we  might  express  any  number  as  an  approximate  power 
of  10. 

436.  By  a  method  of  advanced  algebra,  these  approximate 

values  for  x  have  been  obtained.     For  example,  it  has  been 

found  that 

180  =  io2-2663, 

4500  =  103-6532, 

19600  =  lO*^23,  etc. 

437.  These  exponents  are  called  the  logarithms  of  the  num- 
bers they  produce. 

438.  The  exponent  that  must  be  given  10  in  order  to  produce 
a  required  number,  JV",  is  called  the  logarithm  of  N  to  the  base  10. 

The  expression  10*  =  N 

is  usually  written  x  =  logic  -ZV» 

and  is  read,         M x  is  the  logarithm  of  N  to  the  base  10." 

364 


THE  PARTS  OF  A  LOGARITHM  365 

The  object  and  use  of  logarithms  is  to  simplify  numerical 
work  in  the  processes  of  multiplication,  division,  involution, 
and  evolution. 

439.  Logarithms  to  the  base  10  are  known  as  common  log- 
arithms, and  are  in  universal  use  for  numerical  operations. 

Unless  otherwise  stated,  the  discussions  of  this  and  subsequent  chap- 
ters refer  to  common  logarithms. 

Any  positive  number,  except  unity,  may  be  taken  as  the  base  of  a 
system  of  logarithms. 

A  negative  number  is  not  considered  as  having  a  logarithm. 

THE  PARTS  OF  A  LOGARITHM 

440.  Consider  the  results  in  the  following : 

IO3   =  1000,  log  1000  =  3, 

IO2    =  100,  log  100  =  2, 

101    =io,  log  10     =1, 

10°    =1,  logl       =0, 

10-i  =  .1,  log.i      =_l, 

IO"2  =  .01,  log  .01    =  -  2, 

IO"8  =  .001,  log  .001  =  -  3  ;  etc. 

From  these  results  it  is  evident  that 

(1)  The  common  logarithm  of  a  number  greater  than  1  is 
positive. 

(2)  The  common  logarithm  of  a  number  between  0  and  1  is 
negative. 

»     (3)  The  common  logarithm  of  an  integral  or  a  mixed  number 

between  1  and  10  is  0  +  a  decimal, 
between  10  and  100  is  1  +  a  decimal, 
between  100  and  1000  is  2  +  a  decimal,  etc. 
(4)  The  common  logarithm  of  a  decimal  number 
between  1  and  0.1  is  —  1  -f-  a  decimal, 
between  0.1  and  0.01  is  —  2  +  a  decimal, 
between  0.01  and  0.001  is  —  3  4-  a  decimal,  etc. 


366  LOGARITHMS 

441.  The  integral  part  of  a  logarithm  is  the  characteristic. 
The  decimal  part  is  the  mantissa. 

In  log  352  =  2.5465,  2  is  the  characteristic  and  .5465  is  the  mantissa. 

I.  To  obtain  the  characteristic  of  a  logarithm. 

(a)  When  the  given  number  is  integral  or  mixed. 

442.  By  Art.  440  (3),  the  characteristic  of  the  logarithm  of  a 
number  having  one  digit  to  the  left  of  the  decimal  point  is  0, 
of  a  number  having  two  digits  to  the  left  of  the  decimal  point 
is  1,  and  of  a  number  having  three  digits  to  the  left  of  the 
decimal  point  is  2.     In  general, 

The  characteristic  of  the  logarithm  of  a  number  greater  than 
unity  is  1  less  than  the  number  of  digits  to  the  left  of  the  decimal 
point. 

(b)  When  the  given  number  is  a  decimal. 

443.  By  Art.  440  (4),  the  characteristic  of  the  logarithm  of  a 
decimal  having  no  cipher  between  its  decimal  point  and  its 
first  significant  figure  is  —  1,  of  a  decimal  having  one  cipher 
between  its  decimal  point  and  its  first  significant  figure  is  —  2, 
and  of  a  decimal  having  two  ciphers  between  its  decimal  point 
and  its  first  significant  figure  is  —  3. 

In  order  to  avoid  writing  these  negative  characteristics  —  1, 
—  2,  —  3,  etc.,  it  is  customary  to  consider  that 

-  1  =  9  -  10, 

-  2  =  8  -  10, 

-  3  =  7  -  10,  etc. 

With  the  negative  results  written  in  this  form,  we  have,  in 
general, 

The  characteristic  of  the  logarithm  of  a  number  less  than  unity 
is  obtained  by  subtracting  from  9  the  number  of  ciphers  between 
its  decimal  point  and  the  first  significant  figure,  annexing  — 10 
after  the  mantissa. 


USE  OF  FOUR-PLACE  TABLES  367 

II.   To  obtain  the  mantissa  of  a  logarithm. 

(a)  Important  principle  governing  the  finding  of  all  mantissas. 

444.  It  has  been  computed  that  the  logarithm  of  35,200  is 
4.5465,  and  from  our  discussion  we  have  seen  that  the  logarithm 
of  100  is  2.     We  may  write,  therefore, 

104.5465  =  35200  (1) 

and  102  =  100.  (2) 

These  logarithms  being,  by  definition,  exponents,  we  may 
treat  them  as  such  in  the  following  operation. 
Dividing  (1)  by  (2),  we  have : 

104.5466     35200 


102          100 

From  which, 

102.5465  =  352. 

That  is, 

log  352  =  2.5465. 

Clearly,  therefore,  the  mantissas  of  the  logarithms  of  352 
and  35,200  are  equal. 

In  like  manner,  we  may  show  that 


In  general : 


log  35.2  =  1.5465, 
log  3.52  =  0.5465, 
log  .352  =  9.5465  -  10,  etc. 


If  two  numbers  differ  only  in  the  position  of  their  decimal  points, 
their  logarithms  have  the  same  mantissas. 


THE  USE  OF  THE  FOUR-PLACE  TABLE 

445.   The  table  of  logarithms  is  used  for  two  distinct  and 
opposite  operations. 

(1)  Given  a  number,  to  find  the  corresponding  logarithm. 

(2)  Given  a  logarithm,  to  find  the  corresponding  number. 


368  LOGARITHMS 

I.  To  find  the  logarithm  of  a  given  number. 

446.    (a)  Numbers  having  three  figures. 
Illustrations : 

1.  What  is  the  logarithm  of  247  ? 

On  page  370,  in  the  column  headed  "N,"  we  find  "24,"  the  first  two 
figures  of  the  given  number. 

In  the  same  horizontal  line  with  24,  under  the  heading  corresponding 
to  the  last  figure  of  the  given  number,  "7,"  we  find  the  mantissa,  3927. 

Since  the  given  number  has  three  figures  to  the  left  of  the  decimal 
point,  the  required  characteristic  is  2  (Art.  442). 

Therefore,  log  247  =  2.3927.    Result. 

2.  What  is  the  logarithm  of  .0562  ? 

Opposite  "56"  of  the  "N"  column  (p.  371),  and  under  "2,"  we 
find  the  mantissa  7497. 

Since  the  given  number  is  a  decimal  and  has  one  cipher  between  its 
decimal  point  and  its  first  significant  figure,  we  subtract  1  from  9,  and 
annex  —  10  to  the  mantissa  (Art.  443). 

Therefore,  log  .0562  =  8.7497  -  10.    Result. 

(b)  Numbers  having  two  figures. 
Illustrations : 

1.  What  is  the  logarithm  of  76  ? 

Opposite  "76"  of  the  "N"  column,  and  under  "0,"  we  find  the 
mantissa  8808.    The  characteristic  is  1  (Art.  442). 

Therefore,  log  76  =  1.8808.     Result. 

2.  What  is  the  logarithm  of  .0027  ? 

Opposite  "  27  "  of  the  "  N  "  column,  and  under  "0,"  we  find  the  man- 
tissa 4314.  The  characteristic  is  (9—2),  or  7,  with  —10  annexed 
(Art.  443). 

Therefore,  log  .0027  =  7.4314  -  10.     Result. 


INTERPOLATION  369 

(c)  Numbers  having  onejlgure. 
Illustrations : 

1.  What  is  the  logarithm  of  7  ? 

Opposite  "70"  of  the"N"  column,  and  under  "0,"  we  find  the 
mantissa  8451.    The  characteristic  is  0  (Art.  442). 
Therefore,  log  7  =  .8451.     Result. 

2.  What  is  the  logarithm  of  .00008  ? 

Opposite  "80"  of  the  "N"  column,  and  under  "0,"  we  find  the 
mantissa  9031.  The  characteristic  is  (9  —  4),  or  5,  with  —  10  annexed 
(Art.  443). 

Therefore,  log  .00008  =  5.9031  -  10.    Result. 

447.  It  is  evident  that  the  logarithms  of  numbers  having 
one  or  two  figures  have  mantissas  from  the  "  0  "  column  of  the 
table. 


Exercise  132 

K 

nd  the 

logarithms  of  the  following  numbers : 

l. 

124. 

7.   84.2. 

13.    6.73. 

19. 

.0642. 

2. 

283. 

8.   39.6. 

14.    .829. 

20. 

.0006. 

3. 

589. 

9.   2.85. 

15.    .342. 

21. 

.00016. 

4. 

676. 

10.   6.76. 

16.    .676. 

22. 

.0676. 

5. 

643. 

11.  3.70. 

17.    .037. 

23. 

.00809. 

6. 

540. 

12.   5.89. 

18.    .0681. 

24. 

.00000734. 

INTERPOLATION 

448.  Interpolation  is  based  upon  the  assumption  that  the  dif- 
ferences of  logarithms  are  proportional  to  the  differences  of 
their  corresponding  numbers.  While  the  assumption  is  not 
absolutely  correct,  the  results  obtained  are  exceedingly  close 
approximations.  The  process  is  necessary  in  obtaining  the 
logarithms  of  numbers  having  four  places  by  means  of  the 
four-place  table. 

SOM.  EL.  ALG.  — 21 


370 


LOGARITHMS 


N 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

10 

0000 

0043 

0086 

0128 

0170 

0212 

0253 

0294 

0334 

0374 

11 

0414 

0453 

0492 

0531 

0569 

0607 

0645 

0682 

0719 

0755 

12 

0792 

0828 

0864 

0899 

0934 

0969 

1004 

1038 

1072 

1106 

13 

1139 

1173 

1206 

1239 

1271 

1303 

1335 

1367 

1399 

1430 

14 
15 

1461 

1492 

1523 

1553 

1584 

1614 

1644 

1673 

1703 

1732 

1761 

1790 

1818 

1847 

1875 

1903 

1931 

1959 

1987 

2014 

16 

2041 

2068 

2095 

2122 

2148 

2175 

2201 

2227 

2253 

2279 

17 

2304 

2330 

2355 

2380 

2405 

2430 

2455 

2480 

2504 

2529 

18 

2553 

2577 

2601 

2625 

2648 

2672 

2695 

2718 

2742 

2765 

19 
20 

2788 

2810 

2833 

2856 

2878 

2900 

2923 

2945 

2967 

2989 

3010 

3032 

3054 

3075 

3096 

3118 

3139 

3160 

3181 

3201 

21 

3222 

3243 

3263 

3284 

3304 

3324 

3345 

3365 

3385 

3404 

22 

3424 

3444 

3464 

3483 

3502 

3522 

3541 

3560 

3579 

3598 

23 

3617 

3636 

3655 

3674 

3692 

3711 

3729 

3747 

3766 

3784 

24 
25 

3802 

3820 

3838 

3856 

3874 

3892 

3909 

3927 

3945 

3962 

3979 

3997 

4014 

4031 

4048 

4065 

4082 

4099 

4116 

4133 

26 

4150 

4166 

4183 

.4200 

4216 

4232 

4249 

4265 

4281 

4298 

27 

4314 

4330 

4346 

4362 

4378 

4393 

4409 

4425 

4440 

4456 

28 

4472 

4487 

4502 

4518 

4533 

4548 

4564 

4579 

4594 

4609 

29 
30 

4624 

4639 

4654 

4669 

4683 

4698 

4713 

4728 

4742 

4757 

4771 

4786 

4800 

4814 

4829 

4843 

4857 

4871 

4886 

4900 

31 

4914 

4928 

4942 

4955 

4969 

4983 

4997 

5011 

5024 

5038 

32 

5051 

5065 

5079 

5092 

5105 

5119 

5132 

5145 

5159 

5172 

33 

5185 

5198 

5211 

5224 

5237 

5250 

5263 

5276 

5289 

5302 

34 
35 

5315 

5328 

5340 

5353 

5366 

5378 

5391 

5403 

5416 

5428 

5441 

5453 

5465 

5478 

5490 

5502 

5514 

5527 

5539 

5551 

36 

5563 

5575 

5587 

5599 

5611 

5623 

5635 

5647 

5658 

5670 

37 

5682 

5694 

5705 

5717 

5729 

5740 

5752 

5763 

5775 

5786 

38 

5798 

5809 

5821 

5832 

5843 

5855 

5866 

5877 

5888 

5899 

39 
40 

5911 

5922 

5933 

5944 

5955 

5966 

5977 

5988 

5999 

6010 

6021 

6031 

6042 

6053 

6064 

6075 

6085 

6096 

6107 

6117 

41 

6128 

6138 

6149 

6160 

6170 

6180 

6191 

6201 

6212 

6222 

42 

6232 

6243 

6253 

6263 

6274 

6284 

6294 

6304 

6314 

6325 

43 

6335 

6345 

6355 

6365 

6375 

6385 

6395 

6405 

6415 

6425 

44 
45 

6435 

6444 

6454 

6464 

6474 

6484 

6493 

6503 

6513 

6522 

6532 

6542 

6551 

6561 

6571 

6580 

6590 

6599 

6609 

6618 

46 

6628 

6637 

6646 

6656 

6665 

6675 

6684 

6693 

6702 

6712 

47 

6721 

6730 

6739 

6749 

6758 

6767 

6776 

6785 

6794 

6803 

48 

6812 

6821 

6830 

6839 

6848 

6857 

6866 

6875 

6884 

6893 

49 
50 

6902 

6911 

6920 

6928 

6937 

6946 

6955 

6964 

6972 

6981 

6990 

6998 

7007 

7016 

7024 

7033 

7042 

7050 

7059 

7067 

51 

7076 

7084 

7093 

7101 

7110 

7118 

7126 

7135 

7143 

7152 

52 

7160 

7168 

7177 

7185 

7193 

7202 

7210 

7218 

7226 

7235 

53 

7243 

7251 

7259 

7267 

7275 

7284 

7292 

7300 

7308 

7316 

54 

7324 

7332 

7340 

7348 

7356 

7364 

7372 

7380 

7388 

7396 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

FOUR-PLACE   TABLE 


371 


N 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

55 

7404 

7412 

7419 

7427 

7435 

7443 

7451 

7459 

7466 

7474 

50 

7482 

7490 

7497 

7505 

7513 

7520 

7528 

7536 

7543 

7551 

57 

7559 

7566 

7574 

7582 

7589 

7597 

7604 

7612 

7619 

7627 

58 

7634 

7642 

7649 

7657 

7664 

7672 

7679 

7686 

7694 

7701 

59 
60 

7709 

7716 

7723 

7731 

7738 

7745 

7752 

7760 

7767 

7774 

7782 

7789 

7796 

7803 

7810 

7818 

7825 

7832 

7839 

7846 

61 

7853 

7860 

7868 

7875 

7882 

7889 

7896 

7903 

7910 

7917 

62 

7924 

7931 

7938 

7945 

7952 

7959 

7966 

7973 

7980 

7987 

63 

7993 

8000 

8007 

8014 

8021 

8028 

8035 

8041 

8048 

8055 

64 
65 

8062 

8069 

8075 

8082 

8089 

8096 

8102 

8109 

8116 

8122 

8129 

8136 

8142 

8149 

8156 

8162 

8169 

8176 

8182 

8189 

66 

8195 

8202 

8209 

8215 

8222 

8228 

8235 

8241 

8248 

8254 

67 

8261 

8267 

8274 

8280 

8287 

8293 

8299 

8306 

8312 

8319 

68 

8325 

8331 

8338 

8344 

8351 

8357 

8363 

8370 

8376 

8382 

69 
70 

8388 

8395 

8401 

8407 

8414 

8420 

8426 

8432 

8439 

8445 

8451 

8457 

8463 

8470 

8476 

8482 

8488 

8494 

8500 

8506 

71 

8513 

8519 

8525 

8531 

8537 

8543 

8549 

8555 

8561 

8567 

72 

8573 

8579 

8585 

8591 

8597 

8603 

8609 

8615 

8621 

8627 

73 

8633 

8639 

8645 

8651 

8657 

8663 

8669 

8675 

8681 

8686 

74 
75 

8692 

8698 

8704 

8710 

8716 

8722 

8727 

8733 

8739 

8745 

8751 

8756 

8762 

8768 

8774 

8779 

8785 

8791 

8797 

8802 

76 

8808 

8814 

8820 

8825 

8831 

8837 

8842 

8848 

8854 

8859 

77 

8865 

8871 

8876 

8882 

8887 

8893 

8899 

8904 

8910 

8915 

78 

8921 

8927 

8932 

8938 

8943 

8949 

8954 

8960 

8965 

8971 

79 
80 

8976 

8982 

8987 

8993 

8998 

9004 

9009 

9015 

9020 

9025 

9031 

9036 

9042 

9047 

9053 

9058 

9063 

9069 

9074 

9079 

81 

9085 

9090 

9096 

9101 

9106 

9112 

9117 

9122 

9128 

9133 

82 

9138 

9143 

9149 

9154 

9159 

91(55 

9170 

9175 

9180 

9186 

83 

9191 

9196 

9201 

9206 

9212 

9217 

9222 

9227 

9232 

9238 

84 
85 

9243 

9248 

9253 

9258 

9263 

9269 

9274 

9279 

9284 

9289 

9294 

9299 

9304 

9309 

9315 

9320 

9325 

9330 

9335 

9340 

86 

9345 

9350 

9355 

9360 

9365 

9370 

9375 

9380 

9385 

9390 

87 

9395 

9400 

9405 

9410 

9415 

9420 

9425 

9430 

9435 

9440 

88 

9445 

9450 

9455 

9460 

9465 

9469 

9474 

9479 

9484 

9489 

89 
90 

9494 

9499 

9504 

9509 

9513 

9518 

9523 

9528 

9533 

9538 

9542 

9547 

9552 

9557 

9562 

9566 

9571 

9576 

9581 

9586 

91 

9590 

9595 

9600 

9605 

9609 

9614 

9619 

9624 

9628 

9633 

92 

9638 

9643 

9647 

9652 

9657 

9661 

9666 

9671 

9675 

9680 

93 

9685 

9689 

9694 

9699 

9703 

9708 

9713 

9717 

9722 

9727 

94 
95 

9731 

9736 

9741 

9745 

9750 

9754 

9759 

9763 

9768 

9773 

9777 

9782 

9786 

9791 

9795 

9800 

9805 

9809 

9814 

9818 

96 

9823 

9827 

9832 

9836 

9841 

9845 

9850 

9854 

9859 

9863 

97 

9868 

9872 

9877 

9881 

9886 

9890 

9894 

9899 

9903 

9908 

98 

9912 

9917 

9921 

9926 

9930 

9934 

9939 

9943 

9948 

9952 

99 

9956 

9961 

9965 

9969 

9974 

9978 

9983 

9987 

9991 

9996 

N 

1  o 

1 

2 

3 

4 

5 

6 

7 

8 

9 

372  LOGARITHMS 

Illustrations : 

1.  What  is  the  logarithm  of  6874  ? 

Since  the  table  gives  mantissas  for  numbers  having  but  three  places,  we 
will  consider  that  we  are  seeking  the  logarithm  of  687.4.    (See  Art.  444.) 
From  the  table  we  find 

the  mantissa  of  687  =  .8370 
the  mantissa  of  688  =  .8376 
Difference    =  .0006 
That  is,  an  increase  of  1  in  the  number  causes  an  increase  of  .0006  in 
the  mantissa.     Therefore,  an  increase  of  .4  in  the  number  (for  687.4  is 
.4  greater  than  687)  will  cause  an  approximate  increase  in  the  mantissa 
of  .0006  x  .4  =  .0002. 

(The  fifth  decimal  place  being  disregarded  if  less  than  .5  of  the  fourth.) 
Hence,  for  the  mantissa  of  the  logarithm  of  687.4,  we  have 

.8370  +  .0002  =  .8372. 
Therefore,  with  the  proper  characteristic, 

log  6874  =  3.8372.     Result. 

2.  What  is  the  logarithm  of  23.436? 

Consider  the  number  to  be  234.36. 
From  the  table  we  find 

the  mantissa  of  234  =  .3692 

the  mantissa  of  235  =  .3711 

Difference    =  .0019 

Then,  as  above,  .36  x  .0019  =  .000684, 

or,  approximately,  =  .0007. 

Therefore,  the  mantissa  of  the  logarithm  of 

234.36  =  .3692  +  .0007  =  .3699. 
Whence,  log  23.436  =  1.3699.    Result. 

Exercise  133 

Eind  the  logarithms  of  the  following  numbers : 

1.  2762.  5.   8625.  9.   9824.  13.  68.741. 

2.  3894.  6.   8.642.  10.    .06421.  14.  430.05. 

3.  2007.  7.   726.4.  11.    .005432.  15.  .0006941. 

4.  6492.  8.    54.29.  12.   44.212.  16.  .0004682. 


THE  ANTILOGARITHM  373 

II.    To  find  the  number  corresponding  to  a  given  logarithm. 

449.  The  number  to  which  a  given  logarithm  corresponds 
is  called  its  antilogarithm. 

450.  When  the  mantissa  of  a  given  logarithm  is  found  in 
the  four-place  table,  the  antilogarithm  is  readily  found,  and  if 
the  mantissa  given  is  not  found  in  the  table,  a  close  approxi- 
mation for  the  required  antilogarithm  is  obtained  by  a  process 
of  interpolation. 

Illustrations : 

1.  What  is  the  number  whose  logarithm  is  1.4669  ? 

Look  in  the  table  for  the  mantissa  .4669. 

In  the  same  horizontal  line  with  this  mantissa  and  in  the  "  N  "  column, 
we  find  "29,"  the  first  two  figures  of  the  required  number. 

The  mantissa  is  found  under  the  "3." 

Therefore,  the  required  sequence  of  figures  is  293. 

Now  the  given  logarithm  has  a  characteristic,  1.  Therefore  (Art.  442), 
there  must  be  two  figures  to  the  left  of  the  decimal  point  in  the  required 
number.     Hence,  the  antilogarithm  of  1.4669  =  29.3.     Result. 

2.  What  is  the  antilogarithm  of  7.9112  -10  ? 

The  mantissa,  .9112,  lies  in  the  horizontal  line  with  "  81 "  and  under  "  5." 
The  sequence  of  figures  is,  therefore,  815. 

Since  the  characteristic  is  7  —  10,  the  number  has  two  ciphers  between 
the  decimal  point  and  the  first  significant  figure     (Art.  443). 
Therefore,  the  antilog  of  7.9112—10  =  .00815.     Result. 

3.  What  is  the  antilogarithm  of  2.8828  ? 

The  mantissa,  .8828,  is  not  found  in  the  table,  but  lies  between 
the  mantissa  of  log  763  =  .8825 

and  the  mantissa  of  log  764  =  .8831 

The  difference  of  these  mantissas  =  .0006 
That  is,  an  increase  of  .0006  in  the  mantissas  causes  an  increase  of  1  in 
their  antilogarithms. 

Now  between  the  given  mantissa  and  the  next  lowest  mantissa  there 
exists  a  difference  of 

.8828  -  .8825  =  .0003- 


374  LOGARITHMS 

Therefore,  an  increase  of  .0003  in  the  mantissas  will  cause  an  approxi- 
mate increase  in  the  antilogarithms  of 

.0003  _  5 

.0006""'  ' 
Hence,  .8828  is  the  mantissa  of  the  log  763.5. 

Or,  antilog  2.8828  =  763.5.    Result. 


Exercise  134 

Find  the  antilogarithms  of  the  following  logarithms : 

1.  1.6085.  8.  0.1088.  15.  8.7219-10. 

2.  1.7284.  9.  3.9799.  16.  7.9007-10. 

3.  2.9345.  10.  2.6897.  17.  6.7370-10. 

4.  3.8317.  11.  4.9733.  18.  9.6477-10. 

5.  0.5509.  12.  9.8987-10.  19.  5.8566-10. 

6.  3.5974.  13.  9.7306-10.  20.  9.9995-10. 

7.  2.0095.  14.  8.9099-10.  21.  4.9543-10. 

THE  PROPERTIES  OF  LOGARITHMS 

451.  In  any  system,  the  logarithm  of  1  is  0. 

For,  a0  =  1.  (Art.  241) 

Whence,  log  1=0.  (Art.  438) 

452.  In  any  system,  the  logarithm  of  the  base  is  1. 

For,  a1  =  a. 

Whence,  log  a  =  1.  (Art.  438) 

While,  for  convenience,  the  base  of  the  common  system  is 
used  in  the  following  discussions,  each  theorem  is  a  general 
one  for  any  base,  a,  hence,  for  any  system. 

453.  The  logarithm  of  a  product  is  equal  to  the  sum  of  the 
logarithms  of  its  factors. 


THE  PROPERTIES   OF  LOGARITHMS  375 

Let  10*  =  to  (1)     and     10^  =  n.  (2) 

Multiplying  (1)  by  (2),  \vx+v  =  mn. 

Whence,  log  mn  =  x  +  y.    (3)     (Art.  438) 

From  (1)  and  (2),  x  =  log  to  and  y  =  log  n.  (Art.  438) 

Substituting  in  (3),  log  mn  =  log  to  +  log  n. 

454.    The  logarithm  of  a  quotient  is  equal  to  the  logarithm  of 
tlie  dividend  minus  the  logarithm  of  the  divisor. 
Let  10*  =  m  (1)     and     10*/  =  n.  (2) 


Dividing  (1)  by  (2), 

iv  _  m 
10»  ~~  n 

That  is, 

10*-y  =  Tl. 
n 

Whence, 

log^  =  x-y.                        (3) 
n 

(Art.  438) 

But,  from  (1)  and  (2), 

x  =  log  m  and  y  =  log  n. 

(Art.  438) 

Hence,  in  (3), 

log  —  =  log  m  —  log  n. 
ft 

455.  The  logarithm  of  a  power  of  a  number  is  equal  to  the 
logarithm  of  the  number  multiplied  by  the  exponent  of  the  power. 

Let  10*  =  m.  (1) 

Raising  both  members  of  the  equation  to  the  pth.  power, 

IOp*  =  mP. 
Whence,  log  mP  =  px. 

But,  from  (1),  x  =  log  m,  (Art.  438) 

Hence,    .  log  m"  =p(log  m). 

456.  TJie  logarithm  of  a  root  of  a  number  is  equal  to  the 
logarithm  of  the  number  divided  by  the  index  of  the  root. 

Let  10*  ==  w.  (1) 

Raising  both  members  to  the  power  indicated  by-, 

x  1  0 

10*  =  TO*. 
1 

Whence,  log  mq  =  -.  (Art.  438  ) 

But,  from  (1),  x  =  log  to,  (Art.  438) 

-  _  log  to 
Therefore,  log  mq  —  — - — 


376  LOGARITHMS 

457.   These  demonstrations  may  be  briefly  summarized  as 
follows : 

(1)  To  multiply  numbers,  add  their  logarithms. 

(2)  To  divide  numbers,  subtract  the  logarithm  of  the  divisor 
from  the  logarithm  of  the  dividend. 

(3)  To  raise  a  number  to  a  power,  multiply  the  logarithm  of 
the  number  by  the  exponent  of  the  required  power. 

(4)  To  extract  a  root  of  a  number,  divide  the  logarithm  of  the 
number  by  the  index  of  the  required  root. 

The  antilogarithm  of  a  result  obtained  by  one  of  these  processes 
is  the  required  number. 


THE  COLOGARITHM 

458.  The  cologarithm  of  a  number  is  the  logarithm  of  the 
reciprocal  of  the  number.  By  its  use  an  appreciable  saving 
of  labor  is  made  in  computations  with  logarithms. 


By  definition,    I 

cologiV=log— 

2v 

=  log  1  —  log  N 

(Art.  454) 

=  0- log  .2V 

(Art.  451) 

=  -log2V.      . 

To  avoid  this  negative  form  we  may  write, 

colog  JV=  10  -  log  2V—  10. 

459.   In  general,  to  obtain  the  cologarithm  of  a  number : 

Subtract  the  logarithm  from  10  — 10. 

In  practice  the  subtraction  is  usually  accomplished  by  be- 
ginning at  the  characteristic  and  subtracting  each  figure  from 
9,  excepting  the  last  significant  figure,  which  is  subtracted  from 
10. 


LOGARITHMS   IN   COMPUTATIONS  377 

Illustration : 

Find  the  cologarithm  of  16. 

The  log  of  1  is  0,  which  we  may  write  as  10  —  10. 
Then,  colog  16  =  log  1  —  log  16. 

log   1  =  10  -  10 

log  16  =    1.2041 
Subtracting,  colog  16  =    8.7959  —  10.     Result. 

460.  If  the  characteristic  of  a  logarithm  is  greater  than  10 
but  less  than  20,  we  use  in  like  manner, 

colog  2V  =  20  -  log  N—  20. 
Similarly,  30  —  30,  40  —  40,  etc. ,  may  be  used  if  necessary. 

461.  Negative  factors  in  computations  with  logarithms.  The 
logarithms  of  negative  factors  in  any  group  are  found  without 
regard  to  the  negative  signs,  and  the  result  is  positive  or  nega- 
tive according  as  the  number  of  negative  factors  is  even  or 
odd. 

USE  OP  LOGARITHMS  IN  COMPUTATIONS 

462.  Illustrations  : 

1.  Multiply  6.85  by  37.8. 

log  6.85  =  0.8357 
log  37.8  =  1.5775 
(Art.  453),  2.4132  =  log  of  product, 

antilog  2.4132  =  258.9.    Result. 

2.  Divide  30,400  by  1280. 

log  30400  =  4.4829 
log    1280  =  3.1072 
(Art.  454) ,  1 .  3757  =  log  of  quotient, 

antilog  1.3767  =  23.75.     Result. 

3.  Divide  2640  by  36,900. 

log    2640  =  3.4216 
(Art.  458),  colog  36900  =  5.4330  -  10 

8.8546  -  10  =  log  of  quotient, 
antilog  8.8546  -  10  =  0.07155.    Result. 


378  LOGARITHMS 

4.  What  is  the  value  of  ^846"? 

log  846  =  2.9274. 
(Art.  456)  3)2.9274 

.9758  =  log  of  cube  root, 
antilog  .9758  =  9.458.    Result. 

5.  Find  the  value  of  ^.00276. 

log  .00276  =  7.4409  -  10. 
When  a  negative  logarithm  occurs  in  obtaining  a  root,  the  characteris- 
tic must  be  written  in  such  a  form  that  the  number  subtracted  from  the 
logarithm  shall  be  10  times  the  index  of  the  root.    The  divisor  in  this 
case  being  6,  we  change  the  form  of  the  logarithm  by  adding  50  —  50, 
and  the  subsequent  division  gives  the  required  negative  characteristic, 
log  .00276  =  7.4409  -  10. 
Adding,  50  -  50 

57.4409-60 
(Art.  456)  6)57.4409-60 

9.5735  -  10  =  log  of  sixth  root, 
antilog  9.5735  -  10  =  .3745.    Result. 

4322x27.13 


6.   Find  the  value  of 


35.14 


(Art.  455)  2  log  432  =  2  (2.6355)  =  5.2710 

(Art.  455)  3  log  27 . 1  =  3  (1 .4330)  =4 .2990 

(Arts.  455,  458)  4  colog  35.1  =  4  (8.4547  -10)  =  3.8188  -  10 

3.3888  =  log  of  result, 
antilog  3.3888  =  2448.    Result. 


7.    Simplify  V/2720*  X  V288  X  432*. 
.0687  X  V27.83  x  6242 

\  log  2720     a  |  (3.4346)  =  1.7173 

|  log    288     z=  i(2.4594)  =1.2297 

-flog    432     =|(2.6355)  =3.5140 

7  colog  .068  =  7  (1.1675)  =8.1725 

f  colog  27.8  =  |  (8.5560  -  10)  =  7.8340  -  10 
2  colog  624  =  2  (7.2048  -10)  =  4.4096  -  10 

6.8771 
(Art.  456)  4)6.8771 

1.7193,  log  of  result, 
antilog  1.7193  =  52.40.    Result. 


MISCELLANEOUS  APPLICATIONS  OF  LOGARITHMS     379 


1.  4600  x. 85. 

2.  72x380. 

3.  .28  x. 00012. 

4.  .017  X  .0062. 

5.  4.96  x  58.4. 

6.  .00621  x  .000621 

7.  73400  x  .00811. 

8.  .0293  x  .000602. 

9.  691  x  .0000131. 
624  x  372  x  891 


19 


20. 


21. 


Exercise  135 

e  value  of : 

10. 

1280 -r 

-.0064. 

11. 

68.5  + 

6.12. 

12. 

2.741  - 

-  .00822. 

13. 

.00461 

-  .0931. 

14. 

.07241 

-r-  .3623. 

15. 

(2741 

K  3.623)  - 

-242. 

16. 

(4.625 

X  .5821)  -  2.067. 

17. 

34.74  - 

-  (2.851  x 

4.309). 

18. 

6.904- 

-  (3.676  x 

.00275^ 

.0372 

X  .584  x  . 

30027 

457  x  196  x  583 
43.2  x  3.28  x  .246 
.537x3.41x56.8' 
630  x  2100  x  .007 


23. 


24. 


3.25  X  472  x  6500 

25.  4.522.  28.  ^377. 

26.  2.745.  29.  VWM. 

27.  .02764.  30.  a/,00724. 


.273  x  .00042  x  .0121 
.007  x  .07  x  .7  X  7 
35.7  x  7.14  x  .1428* 
.643  X. 0468x2760 
346  x  .0072  x  .01  ' 

31.  3.207f. 

32.  5.602*. 

33.  .0007544. 


34. 


4 


,00273x5.824x7623 
3.2074x6.423x26.72 


35. 


4 


V.2809  x  ^.003241 
V^962X  ^.000011 


MISCELLANEOUS  APPLICATIONS  OF  LOGARITHMS 

463.   I.    Changing  the  base  of  a  system  of  logarithms. 

Let  a  and  b  represent  the  bases  of  two  systems  of  logarithms,  and  m 
the  number  under  consideration. 


We  are  to  show  that 


logbm 


l0ga& 


380  LOGAKITHMS 

Let  a*  =  m  (1) 

and  bv  =  m.  (2) 

Then,        '  x  =  \ogam  and  y  =  logbm.  (Art.  438) 

From  (1)  and. (2),  &  =  W . 

X 

Extracting  the  yth  root,        a*  =  b. 

Therefore,  log«&  =  -  or  y  =  — — 

V  logab 

That  is,  log6m=^^. 

loga6 

Illustration: 

1.   Find  the  logarithm  of  12  to  the  base  5. 

By  Art.  463,  log612=^  " 


logi05 
= 1.0792 

.6990 
=  1.5439.    Result. 

II.   Equations  involving  logarithms. 

464.  An  equation  in  which  an  unknown  number  appears  as 
an  exponent  is  called  an  exponential  equation.  Thus :  ax  =  b 
is  a  general  form  for  such  equations. 

Illustrations : 

1.  Solve  the  equation,  4X  =  64. 

If,  in  an  equation  in  the  form  of  ax  —  b,  b  is  an  exact  power  of  a,  the 
solution  is  readily  obtained  by  inspection. 

From  4*  =  64, 

we  have  4*  =  43. 

Therefore,  x  =  3.     Result. 

2.  Find  the  value  of  x  in  5*  =  13. 
Using  logarithms,  *  log  5  =  log  13. 


x 


log  5 
1.1139 


.6990 

1.5935.    Result. 


MISCELLANEOUS  APPLICATIONS   OF  LOGARITHMS     381 

3.    Solve  the.  equation  2 s/x  =  VS. 
8/<r         3$ 

2  Vce  =  y/3,     \Jx  = ,     x  =  —  (changing  form  and  squaring) . 

A  4 

Then, 


log  x  =  |  log  3  +  colog  4 

f  log3  =  f  (.4771)  =    .3181 

colog  4=                      9.3979-10 

.     Therefore,                  9.7160  -  10  = 

:  log  Of  X 

x  =  .52.     Result. 

465.  Certain  forms  of  equations  involving  logarithms  may- 
be so  transformed  as  to  give  results  without  the  aid  of  loga- 
rithm tables. 

1.   Find  a  if  logx33  =  5.  2.   Find  x  if  log3  x  =  4. 

By  logarithms,  32  =  x5.  By  logarithms,  x  =  34. 

Whence,  x  =  2.    Result.  Whence,  x  =  81.    Result. 

(That  is,  the  base  of  that  system        (That  is,  the  number,  whose  log 

in  which  the  log  of  32  is  5,  is  2.)  to  the  base  3  is  4,  is  81.) 

III.    Use  of  formulas  for  compound  interest  and  annuities. 

466.  If  P  is  a  given  principal,  n  the  number  of  years  during 
which  the  interest  is  compounded  annually,  and  r  the  given 
rate  per  cent,  the  amount,  A,  can  be  obtained  from  the  formula 

A  =  P(l  +  r)\ 
Illustration : 

1.  Find  the  amount  of  $  1500  for  8  years  at  5  %,  compounded 
annually. 


In  the  formula 

A  =  P(l  +  r)». 

By  substitution, 

A  =  1500  (1.05)8. 

By  logarithms, 

log  A  =  log  1500  +  8  (log  1.05) 

log  1500  =                      3.1761 

8  (log  1.05)  =  8  (.0212)  =     .1696 

log  .4=                      3.3457 

A  =  $2216.50.     Result. 

382 


LOGARITHMS 


467.  An  annuity  is  a  fixed  sum  of  money  payable  yearly, 
or  at  other  fixed  intervals. 

If  the  amount  of  an  annuity  is  represented  by  A,  the  number 
of  yearly  payments  by  %  and  the  rate  of  money  at  the  present 
time  by  r,  the  present  value  of  the  annuity,  P,  can  be  obtained 
from  the  formula 


p=A 


1 


Illustration : 

2.   Find  the  present  value  of  annuity  of  $  900  for  20  years 


at  4%. 

In  the  formula, 

By  logarithms, 
Hence, 

Then, 

By  logarithms, 


r  L        (1  +  r-)2°J     .04  L 


(1.04)' 


20  (log  1.04)  =  20  (.0170) 
(1.04)20  -  2.1879. 

P  =  9oor1 1 

.04  L       2.1879 


-.3400. 


3= 


900  x  1.1879 
.04  x  2.1879  ' 


log  900  =2.9542 
log  1.1879=    .0747 
colog.04      =1.3979 
colog  2.1879  =  9.6600  -10 

4.0868  =  log  P 
P=  $12,211,  approximately.    Result. 


Calculate : 

1.  log,  12. 

2.  log6  28. 

3.  log7  42. 

4.  log1254. 

Solve : 

13.  a;  =  12  VR 

14.  xV5  =  70. 


Exercise  136 

5.  log1560. 

6.  log3256. 

7.  log8  9.6. 

8.  log16.7. 

15.  x-V3=</7. 

16.  12vz  =  19. 


9.  logL6.48. 

10.  loglM.416. 

11.  log.5.875. 

12.  log.9.007. 

17.  2V3^=5V^ 

18.  10  a;"*  =  64. 


MISCELLANEOUS   APPLICATIONS   OF   LOGARITHMS     383 


19.    -v^TaJ2=VI(J.       23.  5Z  =  20.  28.  5X"1  =  35. 

24.  24*  =  100.  29.  16*  =  64-1. 

25.  3.5*  =  170.  30.  (-J )-  =  16. 

21.  5*  =  125.               26  12.4*  =  .124.  31.  25*+2  =  30*"1. 

22.  3- =  729.                27.  3*+1  =  12.  32.  1.62x"1=3.24. 


20.    (3a>)*=</20. 


Given:  log 2  =  .3010,  log 3  =  .4771;  find  logarithms  of  the 
following : 

33.  log  6,  log  18,  log  72,  log  2.88. 

34.  log  3  V2,  log  4  V3,  log  V60,  log  ^045. 

35.  log  270,  logl6f,  log2i  logl22. 

36.  log^0-,  log^a,  log  (2*  -*-  3*  x  6*). 
.Find  x  if 

37.  logx27  =  3.  40.   logx125  =  -3. 

38.  logz16  =  4.  41.   log,  64  =  If 

39.  logx216=3.  42.    log.JV  =  -f 
Find  x  if 

43.  logs  &  =  5.  46.    log49a?  =  —  J. 

44.  log4a;  =  6.  47.    log8»  =  l|. 

45.  log25a;  =  i  48.    log27a;  =  --|. 
What  is  the  expression  for  x  in : 

49.  ax  =  c2.  52.    ax~1  =  c\ 

50.  2ax  =  mwa!.  53.    acx  =  mnx+2. 

51.  4c8  =  oflf.  _  54.    c*m  =  niy+1. 

55.  Find  the  amount  of  $  600  for  10  years  at  4  %  compound 
interest. 

56.  Find  the  amount  of  $  7000  for  15  years  at  4  %  compound 
interest. 

57.  In  what  time  will  $  5000  amount  to  $  7500  at  5  %  com- 
pound interest? 


384  LOGARITHMS 

58.  A  man  buys  an  annuity  of  $  600  for  20  years.  If  money 
is  worth  4  %,  what  amount  should  he  pay  for  it  ? 

59.  What  is  the  cost  of  an  annuity  of  $  1000  per  year  for  10 
years,  money  being  worth  5  %  ? 

60.  Find  the  present  value  of  an  annuity  of  $1200  for  15 
years  at  5%. 

61.  Write  as  a  polynomial,  log  x  (x2  —  l)2(x  +  l)-2. 

62.  Change  2  log  a  -f  log  b  —  log  3  —  \  log  c  to  an  expression 
of  a  single  term. 

63.  Find  an  expression  for  x  in  m**1  =p2. 

64.  If  log  429  =  2.6325,  and  log  430  =2.6335,  find  log  429.3. 
What  is  the  log  .004293? 

65.  If  log  864  =  2.9365,  and  log  8.65  =  0.9370,  find  log 
.08642.     What  is  the  log  864.2  ? 

66.  How  may  you  obtain  log  3,  log  9,  log  243,  and  log  .9  if 
you  know  log  81  ? 

67.  How  many  digits  in  2525  ? 

68.  Find  the  values  of  x  and  y  in  the  equations  3X+V  =  27 
and  S2x~»  =  512. 

69.  If  a,  b,  and  c  are  the  sides  of  any  triangle,  and  s  is  one 
half  their  sum,  the  area  of  the  triangle  may  be  obtained  from 
the  formula,  A  =  Vs  (s  —  a)(s  —  b)(s  —  c).  Write  this  expres- 
sion in  logarithmic  form. 

70.  With  the  formula  of  example  69  find  the  area  of  a 
triangle  whose  sides  are  65  feet,  70  feet,  and  75  feet  respect- 
ively. 

71.  The  diameter  of  a  circle  circumscribing  an  equilateral 
triangle  equals  f  the  altitude  of  the  triangle.  If  h  is  the  alti- 
tude of  an  equilateral  triangle,  the  area  may  be  obtained  from 
the  formula,  A  =  |7i2V3.  Using  logarithms,  find  the  area  of 
the  three  segments  cut  from  a  circle  by  an  equilateral  triangle 
whose  altitude  is  9  feet. 


CHAPTER   XXVIII 
SUPPLEMENTARY  TOPICS 

THE  REMAINDER  THEOREM 

468.  If  any  polynomial  of  the  form,  C^  +  C^cn~l  +  C<p?~2  -\ 

Cn,  be  divided  byx  —  a,  the  remainder  will  be  Qa"  +  C2an~l+  Csan~2 
+  •  •  •  Cn ;  which  expression  differs  from  the  given  expression  in 
that  a  takes  the  place  of  x. 

Proof:  Let  Q  denote  the  quotient  and  B  the  remainder  when 
Cixn  +  C2X11-1  +  Czxn-2  H Cn  is  divided  by  x  -  a. 

Continue  the  division  until  the  remainder  does  not  contain  x. 
Then,  Q(x  -  a)+  B  =  dx*  +  C2xn~l  +  C3xn-8  +  •••  C„. 
Since  this  identity  is  true  for  all  values  of  x,  let  x  =  a. 
Then,  Q(a  -  a)+  B  =  dan  +  C^a"-1  +  C3an-2  +  •••  C„. 
And,  B  =  Cian  +  C2an_1  +  Csan-*  +  •••  C„. 

Application : 

1.  Without  division  obtain  the  remainder  when  7^  +  3^ 
— 13  x  -|-  8  is  divided  by  x  —  2. 

We  have  given,  $e  —  a  =  x  —  2,  hence  a  =  2. 

Hence,  22  =  7  •  2^  +  3  •  22  -  13  -2  +  8  =  60.    Result. 

THE  FACTOR  THEOREM 

469.  If  any  rational  and  integral  expression  containing  x  be- 
comes .  equal  to  0  when  a  is  substituted  for  x,  the  expression  is 
exactly  divisible  by  x  —  a. 

Proof  :  Let  E  be  the  given  expression,  and  let  E  be  divided  by  x  —  a 
until  the  remainder  no  longer  contains  x.  Let  Q  denote  the  quotient  ob- 
tained and  B  the  remainder. 

som.  el.  alg.  —  25  386 


386  SUPPLEMENTARY  TOPICS 

Then,  E  =  Q(x  -a)  +  B. 

This  identity  being  true  for  all  values  of  x,  we  may  assume  that  x  =  a. 
By  the  hypothesis  the  substitution  of  a  for  x  makes  E  equal  to  0. 

Therefore,     0  =  E(a  -  a)  +  B,    0  =  0  +  B,     B  =  0. 

Therefore,  the  remainder  being  0,  the  given  expression  is  exactly 
divisible  by  x  —  a. 

Or,  x  —  a  is  a  factor  of  the  given  expression,  E. 

Illustrations : 

1.  Factor  Xs- 19  x  +  16. 

By  trial  we  find  that  the  expression,  x8— 12  a; +16,  equals  0  when  x  =  2. 

Therefore,   if  x  =  2,  x  —  a  =  x  —  2,  and  x  —  2  is  a  factor. 

Then  (z3  -  12  x  +  16)  -h  (x  -  2)  =  x2  +  2  g  -  8. 

The  factors  of  x2  +  2  x  —  8  are  found  to  be  (x  +  4)  and  (x  —  2). 

Therefore,  x3  -  12  x  +  1 6  =  (x  -  2)(x  +  4)(x  -  2).     Result. 

2.  Factor  x*+9x*  +  23x  -f- 15. 

(In  an  expression  whose  signs  are  all  plas,  it  is  evident  that  no  positive 
number  can  be  found  the  substitution  of  which  will  make  the  expression 
equal  to  0.) 

By  trial  we  find  that  if  x  =  —  1,  the  expression  becomes  0. 

Hence,  if  x  =  —  1,  x  —  a=[x  —  (  —  1)]  =  x  +  1,  a  factor  required. 

Then,  (x8  +  9x2  +  23 x  +  15)  +  (z  +  1)  =  x2  +  8x  +  16. 

Furthermore,  x2  +  8  x  +  15  =  (x  +  3)  (x  +  5). 

Therefore,  x8  +  9  x2  +  23  x  +  15  =  (x  +  1)  (x  +  3)  (x  +  5) .    Result. 

Exercise  137 

Without  division  show  that 

1.  a3  +  3  a2  4-  3  a  +  2  is  divisible  by  a  +  2. 

2.  ^-8^  +  24^-3205  +  16  is  divisible  by  x  —  2. 

3.  c4-c8-8c2  +  9c-9is  divisible  by  c  +  3. 

4.  27a3  +  9a2-3a-10  is  divisible  by  3a-2. 

5.  m5  —  5  m4  +  9  m3  —  6  m2  —  m  +  2  is  divisible  by  m  —  2. 


DIVISORS  OF  BINOMIALS  387 

Factor : 

6.  c8  +  c-2.  12.  c8  +  c2-5c  +  3. 

7.  a?-7x  +  6.  13.  rf  +  Zrf-bx-Z. 

8.  m3-8m  +  3.  14.  m3  +  5m2-13m  +  7. 

9.  a8  +  3a2  +  7a  +  5.  15.  2 2/3  +  3 1/2  —  3 y  —  2. 

10.  ^  +  4^  +  5^  +  2.  16.   3a3-10a2-f  4a  +  8. 

11.  m3-2m2-7m-4.  17.   x^-x^-Sx2 -\-5x-~2. 

THE  THEORY  OF  DIVISORS  OF  BINOMIALS 

470.  The  following  proofs  depend  directly  upon  the  prin- 
ciples established  in  Arts.  468  and  469. 

If  n  is  a  positive  integer,  we  may  establish  as  follows  the 
divisibility  of  the  binomials,  xn  —  yn  and  af  -j-  yn,  by  the  bino- 
mials, x  —  y  and  x-^-y. 

I.  xn  —  yn  is  always  divisible  by  x  —  y. 
For,  if  y  is  substituted  for  x  in  xn  —  yn,  we  have 

3jn  _  y»  =  yn  __  yn  —  Q^ 

Therefore,  x  —  y  is  always  a  divisor  of  ccn  —  yn. 

II.  £cn  —  ?/n  is  divisible  byx  +  yifn  is  even. 
For,  if  —  y  is  substituted  for  xm  xn  —  yn,  we  have 

xn  —  yn=(—y)n  —  yn  =  yn  —  yn  =  0. 
Therefore,  x  +  y  is  a  divisor  of  xn  —  yn  when  n  is  even. 

III.  a?  +  yn  is  never  divisible  by  x  —  y. 
For,  if  y  is  substituted  for  x  in  ccn  -f  y»,  we  have 

xn  _j_  yn  —  yn  .J.  yn  —  2  y«. 

which  result  does  not  reduce  to  0  by  the  substitution. 
Therefore,  x  —  y  is  never  a  divisor  of  xn  +  yn. 

IV.  xn  -f-  ?/w  ?'s  divisible  byx  +  yifnis  odd. 
For,  if  —  y  is  substituted  for  x  in  sen  4-  2/*,  we  have 

£«  _j_  ^n  _  (_  y)n  _|_  yn  _  0. 

Therefore,  x  +  2/  is  a  divisor  of  #  +  ?/  when  n  is  odd. 


388  SUPPLEMENTARY  TOPICS 

These  results  may  be  summarized  with  the  corresponding 
quotients  as  follows : 


— — IC  ss  x"-1  +  xn~2y  +  xro-%2  -|-  . .  •  +  w"-1 
x-y 

xn  —  yn  _  ^n_1  _  xn-2y  _j_  xn~8y2  —  •  • .  —  y*-1 

x  +  y 
x  —  y 
x  +  y 


when  n  is  even. 


>,  when  n  is  odd. 


Illustrations  : 

1.  Divide  a?°  +  yw  by  tf  +  tf. 

We  may  write  x10  +  y™  =  (x2)6  +  (y2)5. 

Then,  (x2)6  +  (y2)6  is  divisible  by  a2  +  y2.     (Art.  470,  IV.) 

Whence, 

(S2)6+y)6=(a.2)4  _  (a.2)3(y2)  +  (^)2(2/2)2  _  (^)(y«)8  +  (y2)4 

a;-*  +  y-* 

=  x8  —  xV  -I-  x4*/4  -  xV  +  f.    Result. 

2.  Divide  64  a12  -  tf  by  2  a?  -  */. 

We  may  write  64  x12  —  y*  =  (2  x2)6  —  y*. 

Then, 

(2  x2)6  -  y6  _      ^  +       ^  +  ^  ^^  +  ^  ^g  +  ^  x2)y4  +  y6 
2^  —  y 

=  32  x10  +  16  x8y  +  8  xfy2  +  4  x4y3  +  2  xV  +  y5.    Result. 


Exercise  138 
Divide : 

1.  a?  +  32y5byx  +  2y. 

2.  a?7-128bya;-2. 

3.  32^  +  243  by  2o;  +  3. 

4.  ^-729  by  --S. 

x6  x 

5.  What  are  the  exact  binomial  divisors  of  81  —  a*  ? 


HIGHEST  COMMON  FACTOR  389 

6.  What  are  the  exact  binomial  divisors  of  x10  —  y10  ? 

7.  What  are  the  exact  binomial  divisors  of  x2  —  64  c6  ? 

8.  Obtain  the  three  factors  of  x9  —  y9. 

9.  Obtain  the  four  factors  of  Xs  —  y8. 

10.  .  Obtain  the  six  factors  of  xu  —  4096. 

11.  Obtain  the  ten  factors  of  a12  —  64  a6. 

THE  HIGHEST  COMMON  FACTOR  OF  EXPRESSIONS  NOT  READILY 
FACTORED 

471.  The  method  of  obtaining  the  highest  common  factor  of 
two  expressions  that  cannot  be  factored  by  inspection  is  analo- 
gous to  the  process  of  division  used  in  arithmetic  for  obtaining 
the  greatest  common  divisor  of  two  numbers. 

The  principles  involved  in  the  algebraic  process  are  estab- 
lished as  follows : 

Let  A  and  B  represent  two  expressions,  both  arranged  in  the  descend- 
ing powers  of  some  common  letter ;  and  let  the  degree  of  A  be  not  higher 
than  the  degree  of  B.  Let  B  be  divided  by  A,  the  quotient  being  Q  and 
the  remainder  B. 

Thus,  A)B    (Q 

A$ 

B 

It  follows,  therefore,  that     B  =  AQ  +  B,  (1) 

and  B  =  B-AQ.  (2) 

Now  a  factor  of  each  of  the  terms  of  an  expression  is  a  factor 
of  the  expression. 

We  make,  therefore,  three  important  statements : 

(1)  Any  common  factor  of  A  and  R  in  (1)  is  a  factor  of 
AQ  +  E,  hence  a  factor  of  B.     Or, 

A  common  factor  of  A  and  R  is  a  common  factor  of  B  and  A. 

(2)  Any  common  factor  of  B  and  A  in  (2)  is  a  factor  of 
B—  AQ,  hence  a  factor  of  R.     Or, 

A  common  factor  of  B  and  A  is  a  common  factor  of  A  and  R. 


390 


SUPPLEMENTARY   TOPICS 


(3)  Hence,  the  common  factors  of  B  and  A  are  the  same  as 
the  common  factors  of  A  and  R.     Or, 

The  H.  G.  F.  of  A  and  B  is  the  H.  G.  F.  of  A  and  R. 

Each  succeeding  step  of  the  process  may  be  proved  in  a 
similar  manner. 

If,  at  any  step,  the  remainder  becomes  0,  the  divisor  is  a 
factor  of  the  corresponding  dividend  and  is,  therefore,  the 
H.  G.  F.  of  itself  and  the  corresponding  dividend.     Or, 

The  last  exact  divisor  is  the  required  highest  common  factor. 

472.  We  may  multiply  or  divide  either  given  expression,  or  may 
multiply  or  divide  any  resulting  expression,  by  a  monomial  that 
is  not  a  common  factor  of  both  expressions. 

For  the  process  refers  to  polynomial  expressions  only,  and 
the  introduction  or  the  rejection  of  monomial  factors  not  com- 
mon to  both  expressions  cannot  affect  the  common  polynomial 
expression  sought. 

Such  factors  are  introduced  or  rejected  merely  for  con- 
venience, as  will  be  shown  in  the  following  illustrations. 

In  finding  the  highest  common  factor  of  three  or  more  expres- 
sions, as  A,  B,  and  G,  we  first  find  F1}  the  highest  common 
factor  of  A  and  B.  It  remains  to  find  the  highest  common 
factor  of  F1  and  G.  Let  this  result  be  F2.  Then  F2  is  the 
required  highest  common  factor  of  A,  B,  and  G. 

Illustrations : 

Find  the  H. C.F.  of  2  a*-a*+x-6  and  4  a*+2  a^-lO  x-3. 

The  following  arrangement  is  universally  accepted  as  most  satisfactory  : 


2 as8—      x2+      x-    6 
2 

4x3+    2  a;2  -10  x  -    3 
4  a;8  -    2  x'1  +    2  x  -  12 

X 

4  a;8-    2x2+    2  a; -12 
4  a8  -  12  a;2  +    9  a; 

+    4  x2  -  12  x  +    9 
5 

5x 

+  10  x2-    7  a; -12 
+  10  a:2  -  15  x 

20  xa  -  60  x  +  45 

20  a;2 -14  a; -24 

-23)- 46  a; +  69 

H.C.  F.    2x-   3 

4 

+    8  a; -12 
+   8  a; -12 

HIGHEST  COMMON  FACTOR  391 

Explanation : 

1.  Dividing  4  a;3  +  2  x2  —  10  x  -  3  by  2x*  —  x2  +  x  —  6,  we  obtain  a 
remainder,  4  x2  —  12  x  +  9,  an  expression  lower  in  degree  than  the  divisor 
just  read. 

2.  Dividing  2x3-  x2  +  x  —  6  by  4  oj2  —  12  x  +  9,  we  obtain  a  remainder 
having  such  coefficients  that  the  next  succeeding  division  does  not  reduce 
the  degree  of  the  remainder.  To  obtain  a  division  that  will  reduce  the 
degree  of  the  remainder  we  multiply  4  x2  —  12  *  +  9  by  5,  and  the  product, 
20  x2  —  60  x  +  45,  contains  10  x2  —  7  x  —  12,  with  a  remainder  as  desired, 
—  46  se  +  69.  From  this  remainder  we  may  discard  the  factor,  —23, 
since  this  factor  is  not  common  to  both  expressions  at  this  point.  • 

3.  The  remainder,  2  x  —  3,  divides  10  x2  —  7  x  —  12  exactly. 
Therefore,  2  x  —  3  is  the  required  H.  C.  F. 

It  may  be  noted  that  the  process  might  have  been  completed  by  multi- 
plying 10  x2  —  7  x  —  12  by  2,  the  resulting  product  being  used  as  a  divi- 
dend with  4  x2  —  12  x  +  9  as  a  divisor.  Division  either  way  is  possible 
when  both  expressions  are  like  in  degree. 


Exercise  139 

Find  the  H.C.F.  of: 

1.  ^-3^  +  4a-2and^-r-o2-4a;  +  2. 

2.  a3-2a2-a  +  2anda3-2a2  +  3a-2. 

3.  c3  +  5c2  +  7c-f-3andc3  +  c2-5c  +  3. 

4.  ^_^_5a;_|_2  and  a^  +  4 »2 -f  3 a; - 2. 

5.  m3  +  3m2-f-5ra  +  3  and  m3  +  2m2-2m-3. 

6.  a3-5a2  +  7a-3anda3-a2-7a  +  3. 

7.  c4  +  c2  +  2candc4-c3-3c2-c. 

8.  2a6  +  4ari  +  6a4-f-4arJ  +  2a^  and  4a5- 4 a3 -8^-40;. 

9.  m3  +  ra2  —  m  + 15  and  m3  +  6  m2  +  5  m  — 12. 

10.  2c3  +  5c2  +  5c  +  6and3c3  +  5c2-c  +  2. 

11.  6a;4  +  24(»3  +  30a!2-f-36a;and3a;4-f-6a53-12ic2-9a;. 

12.  2  a4  +  a3  +  3  a2  +  a  +  2  and  a4  +  2  a3  +4  a2  +  3  a  +  2. 

13.  c3-3c2  +  5c-3and2c4-3c3-3c2  +  10c-6. 

14.  2»4+a^-4^  +  3a;-2and3a;4  +  4^-3ar2-4. 


392  SUPPLEMENTARY  TOPICS 

15.  4a4 -4  a3- 5  a2  +  6a- land  6a4-5a3-6a2  + 3a +  2. 

16.  I m5-3m4-2m3+5m2-2m and 3m6-7m5+  6m4- 3m3 

+m2. 

Reduce  the  following  to  lowest  terms : 

1?    m3-4m2  +  2m  +  l,  20       c3-c2-5c-3 


m3-2m2  +  3ra-2  c3_4c2_nc_6 

3-5a  +  7a2  +  3a3   ,  gl     3  a3 +  14  or2 -5  a -56 


6-7a  +  3a2  +  2a3  6a3  + 10  a2  +  17  a +  88 

10  4a2  +  9a-9  00      2m4-2m2  +  m-l 


4a4-f-10a3-7a2  +  9  m*_m3+2m2-m-l 

Find  the  H.  C.  F.  of: 

23.  a3-2a  +  l,  a3-2a2  +  l,  and  a3-2  a2  +  2a-l. 

24.  a3  +  4  a2  +5  x+2,  ^+3^+4^+4,  and  ^+3^+3  a +2. 

25.  2a?4-2af'4-4a;2-4ic,  3 a;4 -f-  9x2-12a;,  and  4a4-8a;3 

-20«2  +  24a-. 

26.  a4-4a2-a  +  2,  a5-  3a3+  3a2-l,  and  a4-f-3a-2. 

THE  LOWEST  COMMON  MULTIPLE  OP  EXPRESSIONS  NOT  READILY 
FACTORED 

473.  In  finding  the  lowest  common  multiple  of  two  ex- 
pressions not  readily  factored  we  first  find  the  highest  common 
factor  by  Art.  471,  after  which  the  process  is  established  by  the 
following. 

Let  A  and  B  represent  any  two  expressions  whose  highest  common 
factor  we  find  to  be  H.     Dividing  both  A  and  B  by  H,  we  obtain 

^  =  a  and  ^  =  6. 
II  H 

Then,                               A-Hxa,  (1) 
B  =  Hxb.  (2) 
Now,  since  H  is  the  highest  common  factor  of  A  and  2?,  the  two  quo- 
tients, a  and  6,  can  have  no  common  factor.  Hence,  for  the  lowest  com- 
mon multiple  of  A  and  B  we  write 

L.C.M.=Rxaxb.  (3) 
Writing  (3)  thus,  L.  C.  M.=5ax  6. 


LOWEST  COMMON  MULTIPLE  393 

Multiplying  and  dividing  by  J9", 

L.C.M.  -Hax^- 
H 

Substituting  Ha  =  A  from  (1),  and  Hb  —  B  from  (2), 

L.C.  M.  =  ^x— . 
H 

In  general,  therefore: 

To  find  the  lowest  common  multiple  of  two  expressions,  divide 
one  of  the  expressions  by  their  highest  common  factor ,  and  multi- 
ply the  other  expression  by  the  quotient. 

The  product  of  two  expressions  is  equal  to  the  product  of  their 
highest  common  factor  by  their  lowest  common  multiple.     For, 

Multiplying  (1)  by  (2),  we  obtain  AB  =  Hx  ax  Hxb 

=  H(Hab). 

From  (3)  L.  C.  M.  =  Hab.    Hence,  AB  =  H(L.  C.  M.). 

In  finding  the  lowest  common  multiple  of  three  or  more  expressions  as 
A,  B,  and  C,  we  first  find  L\,  the  lowest  common  multiple  of  A  and  B. 
It  remains  to  find  the  lowest  common  multiple  of  L\  and  C.  Let  this 
result  be'-L2.  Then  L2  is  the  required  lowest  common  multiple  of  A,  B, 
and  G. 

Illustrations : 

1.   Find  the  L.  C.  M.  of : 

2  ra3  —  ra2  +  m  —  6  and  4  m3  +  2  ra2  — 10  m  —  3. 

By  the  method  of  division  we  find  the  H.C.F.=2m-3. 

Then,  L.  C.  M.  =  (*™*-m*  +  m-Q\  (4  ms  +  2  w2  _  10  m  _  3) 

V  2i»-3  )K  J 

=  (m?  +  m  +  2)(4m8  +  2m2  -  10m  -  3).     Result. 

Exercise  140 
Find  the  L.  C.  M.  of: 

1.  ^  +  5^  +  705  +  3  and<B8  +  <B2-3a>  +  9. 

2.  a3  +  a2-10a  +  8  and  a3-4a2  +  9a- 10. 

3.  2m3  +  5m2-f-2m  —  1  and  3  m3  +  ra2  —  ra  + 1. 

4.  4c3-8c2-3c  +  9and  6c3  +  c2-19c  +  6. 

5.  4ar5-7a  +  3and2;c3  +  ar2-3a;  +  l. 


394  SUPPLEMENTARY   TOPICS 

6.  2a3-7a2-16a  +  5anda3-9a2  +  18a  +  10. 

7.  c3  +  5c2-c-5  and  c4-c3-f-c2  +  c-2. 

8.  a^  +  2^-a;  +  6and2arJ-f-llarJ  +  16a;  +  3. 

9.  m4-3m2  +  l  andm3  +  2m2-4m-3. 

10.  4^-2a;2  +  6a  +  4and6a;4  +  9x3-l5ar2-9a. 

11.  2a4  +  3a3  +  4a2  +  2a  +  l  and  3a4-f-2a3  +  a2-2a-l. 

12.  m5  -  3  m4  +  3  m3-ll  ra2+6  m  and  2  m4-18  ra2+8  m-24. 

13.  4a4-a36  +  2a262  +  2a63  +  &4and3a4-5a3&+9a262 

-6a63  +  464. 

14.  a^+2a;24-2a;4-l,  ^+2^+305+2,  and  a3-^  ^+4^+3. 

15.  4  m3  —  4  m2  4-  3  m  —  1,  6  m3  —  m2  -f-  m  —  1,  and  8  m3  —  2  m2 

+  m-l. 

THE  CUBE  ROOT  OF  POLYNOMIALS 

474.  If  a  binomial,  (a+6),  is  cubed,  we  obtain  (a3+3a26 
-f3a&2-f-&3).  We  have  in  the  following  process  a  method  for 
extracting  the  cube  root,  (a  +  b),  of  the  given  cube,  (a3  +  3  o?b 
+  3a&2  +  63). 

a3 + 3  a2b + 3  ab2 + 68  la  +  6    Cube  Root. 

Trial  Divisor    Complete  Divisor    a? 

3a2  (3a2  +  3ab  +  b2) 


b 


+3a26+3a&2+&8 
+  3a2&+3a62+68 


The  first  term  of  the  root,  a,  is  the  cube  root  of  the  first  term  of  the 
given  expression,  a3.  Subtracting  a3  from  the  given  expression,  the  re- 
mainder, 3  a2b  +  3  ab2  +  ft3,  results. 

The  second  term  of  the  root,  b,  is  obtained  by  dividing  the  first  term  of 
the  remainder,  3  a2b,  by  three  times  the  square  of  a.  This  divisor,  3  a2, 
is  the  Trial  Divisor. 

The  complete  divisor,  (3  a2  +  3  ab  +  b2) ,  consists  of  (the  trial  divisor) 
+  3  (the  trial  divisor)  (the  last  quotient  obtained)  +  (the  square  of  the  last 
quotient). 

The  product  of  the  complete  divisor  by  the  last  quotient  obtained, 
(3  a2  +  3  ab  +  b2)b,  completes  the  cube. 

The  process  may  be  repeated  in  the  same  order  with  any  polynomial 
perfect  cube  whose  root  has  more  than  two  terms. 


CUBE   ROOT  OF  POLYNOMIALS 


395 


Illustration  : 

1.   Find  the  cube  root  of 

8  a6  -36  a5  +  102  a4  -171  a3  +  204  a2  -144a  +  64. 

The  polynomial  is  first  arranged  in  descending  order. 

|2q2-3q  +  4 
8a6-36a5+102a*-171a3+204«2_i44a+64 
8«s 


3(2a2)2=12a4 
3(2  a2)  (-3  a)  =        -18  a9 

(-3q)2= +9a' 


(12a*-18a3+9«2)(-3q): 


-36  a*  +  102  a*- 171  a3 
-36  a5  +  54  a*- 27  a3 


3(2a2-3a)2  =  12a4-36a3+27a2 
3(2a2-3a)(4)  =  24a2-36a 

42=  +16 


(12  a*-  36  a3 +51  a2-  36  a +16)  (4) 


48  a4- 144  a3 + 204  a2- 144  a +64 


48  a4 -144  a3 +204  a2 -144  a +  64 


475.  The  distribution  of  the  volume  of  a  cubic  solid  whose 
edge  is  a  +  6,  and  the  relation  of  the  separate  portions  to  the 
separate  terms  of  the  polynomial  representing  its  cube,  can 
be  readily  seen  from  the  following  illustrations.  The  planes 
cutting  the  cube  pass  at  right  angles  to  each  other  and  parallel 
to  the  faces  of  the  cube ;  each  at  a  distance  of  b  units  from  the 
faces  of  the  cube. 


I  n  m  is? 

In  the  figure  let  the  edge  of  the  cube  in  (I)  be  a  and  the 
edge  of  the  complete  cube  in  (IV)  be  a  +  6. 

I.  a3. 

II.  a3  +  3a26. 

III.  a3  +  3a26  +  3a62. 

IV.  a3  +  3a26  +  3a62+63. 


396  SUPPLEMENTARY   TOPICS 

„  M 

THE  CUBE  ROOT  OF  ARITHMETICAL  NUMBERS 

476.  It  will  be  observed  that 

l8  =           1  A  number  of  one  figure  has  not  more  than  three 

9s  as       729  figures  in  its  cube. 

108  =     1000  A  number  of  two  figures  has  not  more  than  six 

998  =  970299,  etc.  figures  in  its  cube. 

Conversely,  therefore: 

If  an  integral  cube  has  three  figures,  its  cube  root  has  one  figure. 

If  an  integral  cube  has  six  figures,  its  cube  root  has  two  figures,  etc. 

Hence : 

477.  Separate  any  integral  number  into  groups  of  three  figures 
each,  beginning  at  the  right,  and  the  number  of  groups  obtained 
is  the  same  as  the  number  of  integral  figures  in  its  cube  root. 

It  is  to  be  noted  that  the  process  of  cube  root  of  numbers  as 
given  in.  arithmetical  practice  is  based  directly  upon  the  al- 
gebraic principles  learned  in  Art.  474. 

Illustration : 

1.   Find  the  cube  root  of  421875. 

Separating  the  number  into  groups  of  three  figures  each,  we  have  42i875. 

42J875|70  +  5  =  75.     Result. 
a8  =  708  =  343000    a  +  b 


3(a)2  =     3(70)2  =  14700 

3(a)(6)  =3(70)  (5)=    1050 

(6)2  -         (5)2  =       25 

(15775)  (5)  = 


'8875 


78875 


The  analogy  between  the  formation  of  the  cube  of  (a  +  6)  and  the  cube 
of  75,  or  (70  +  5),  may  be  seen  in  the  following : 

(a  +  b)s  =  a3  +  3a26 .+  3«62  +  68 
(70  +  5)8  =  708  +  3(70)2(5)  +  3(70)  (5)2  +  68 
=  343000  +  73500  +  5250  +  125 
=  421876. 


CUBE   ROOT  OF  ARITHMETICAL  NUMBERS 


397 


2^Find  the  cube  root  of  12812904. 

In  practice  the  process  is  usually  abbreviated  as  follows: 

.    12812904|234.     Result. 


2»  = 

8 

3(20)2  _ 

1200 

4812 

3(20)  (3)  = 

180 

32  = 

9 

3(1389)  = 

4167 

3(230)2  = 
3(230) (4)  = 
42  = 


158700 
2760 
16 
4(161476) 


645904 


645904. 


Exercise  141 

Find  the  cube  root  of : 

1.  a3  +  6a2  +  12a+8. 

2.  27  c3  -  54c2  +  36  c -8. 

3.  8a3-36a2  +  54a-27. 

4.  27  a6-  135  a4 +  225  a2  -125. 

5.  a6  +  6a5  +  15a4  +  20a3+15a2  +  6a  +  l. 

6.  ^-6^  +  21a4-44a3  +  63a2-54a-f27. 

7.  a6  +  9a5  +  21a4-9a3-42a2  +  36a-8. 

8.  102a4  +  204  x2-  171a3  -144a +  64  -36a5  +  $x«. 

9.  27m10  -  35 m6  +  12m2  -  8  +  30m4  -  45m8  +  27m12. 
10.  c9-3c8  +  6c7-4c6  +  6c4-2c3  +  3c  +  l. 

11.  74088.  16.  3112.136. 

12.  389017.  17.  1470612$. 

13.  658503.  18.  48.228544. 

14.  912673.  19.  .559476224. 

15.  1953125.  20.  .000138991832. 


398  GENERAL   REVIEW 

GENERAL  REVIEW  -      | 

Exercise  142 

1.   Find  the  numerical  value  of  81"*,  16^",  (i)~4,  5°,  25~t 
Define  briefly  the  law  that  governs  each  reduction. 


2.    Solve   x  +  V.25  jc  +  0.06  =  0. 


3.  Form   the   equation   whose   roots   shall   be  a  —  V  —  1 
and  a  +  V—1. 

4.  Solve  for  x :  2(4  -  3a)^  -  12  =  s& 

5.  Factor  125  ar6-Va7. 

6.  Solve  a?  -f  10  aT  +7  =  0,  and  test  the  roots  obtained. 

7.  By  inspection   determine  the   nature   of   the   roots  of 
Sx2-7x=  -5. 

8.  Solve  4  x2  —  4  ax  +  a2  —  c2  =  0. 

9.  Solve  and  verify  both   solutions   of   (x  +  T)\x  —  2)  = 
a,(a>-3)(&-l)-4. 

10.  Solve  ?— -  =  a?(aj  +  2)  +  1. 

0*  -  !) 

11.  Find  the  square  root  of 

g*-s»  +  l      6(0* +  1)  15a?         20  . 

(a^  +  1)-1  a;-1  (a^  +  l)"1 

12.  Find  the  value  of  m  for  which  the  equation,  3x2  —  6x  + 
m  =  0,  has  equal  roots. 

13.  Factor  2[2  a2  -  (x  -  a)  -  2  a(2 x  -  a)]. 

14.  Write  the  8th  term  of  the  quotient  of  xl5—y15  divided 
by  x-y. 

15.  Solve  x*  +  2xi-  3  =  0. 

16.  Solve    *-2m+     3m     =    *  +  «  ■ 

c  +  d       a?  -j-  2m      a?  +  2m 

17.  Find  the  five  roots  ofar5-4arJ-a?2  +  4  =  0. 


GENERAL  REVIEW  399 

18.  Factor  15  x15  —  15. 

19.  Solve  x(x  ~y)  —  2,  (x  +  y) 2  =  9. 

20.  Form   that   quadratic  equation   whose    roots   shall  be  * 
-  i  and  -  f 

21.  Obtain  the  factors  of  x2  —  5#  +  2  by  solving  the  equa- 
tion, x2  —  5  x  +  2  =  0. 

22.  Divide  16  into  3  parts  in  geometrical  progression,  so 
that  the  sum  of  the  1st  and  2d  shall  be  to  the  difference  of  the 
2d  and  3d  parts  as  3  :  2. 

23.  Find  log  of  7,  given  log  5  =  .6990  and  log  14  =  1.1461. 

24.  Solve  and  test  the  solutions  of 

V2  m  +  x  —  V2  m  —  x  —  V2#  =  0. 

,.-2 


25.    Factor  4  + 4i?  +  m  + 


mx~ 


x  p~x 

26.  Form   the    quadratic   equation    whose    roots    shall   be 
l+V^andl-V^T. 

27.  Solve  x2  +  xy  +  y2  =  7  x, 

x2  —  xy-{-y2  —  Sx. 

28.  What  is  the  sum  of  the  first  n  numbers  divisible  by  5  ? 


„  ^.(H+fXf-M)-*. 


30.  Solve  2s"1  x  2*  =  40. 

31.  Givenl+V«:l  +  3V^  =  2:5;  find  a. 

32.  Factor  ar4  -  2  -f  a4. 

33.  If  7i  is  an  odd  integer,  which  of  the  following  indicated 
divisions  are  possible  ? 

xn  +  yn      xF+y"     xn  —  yn      xn  —  yn 
x  +  y  '     x  —  y  '     x  +  i/  '      #  — ?/ 

34.  Form  the  quadratic  equation  whose  roots  are  1  -+-  a  yi 
and  1  —  a  V#. 


400  GENERAL  REVIEW 

35.  Under  what  condition  will  the  roots  of  ax2  -+-  bx  -f-  c  =  0 
be  imaginary  ?     Prove  your  answer. 

36.  If  the  length  and  breadth  of  a  certain  rectangle  are 
each  increased  by  2  rods,  the  area  will  become  48  sq.  rd. ;  but 
if  each  dimension  were  decreased  by  2  rods,  the  area  would 
become  8  sq.  rd.     Find  the  dimensions  of  the  rectangle. 

37.  Factor  x*  +  2x2-5x-6. 

38.  Write  in  logarithmic  form  27x  =  81,  and  find  x. 

39.  Solve  for  n :  -^ 2  c  = 


n  n  —  1 

40.  What  is  the  price  of  candles  per  dozen  when  3  less  for 
36  cents  raises  the  price  12  cents  per  dozen  ? 

41.  Show  that  the  product  of  the  roots  of  x2  —  5x  —  2  =  0 
is  —4. 

42.  Prove  that  — -,  3,  — - — ,  are  in  geometrical  progression, 

and  find  the  sum  of  10  terms. 

43.  If  m  :  x  =  n  :  y,  show  that 

mx  —  ny  :  mx -\-ny  =  m2—n2:m2  +  n2. 

44.  Find  an  expression  for  the  nth  odd  number,  and  illus- 
trate your  answer  by  a  numerical  substitution. 

45.  Find  n  in  the  formula  I  =  arn~l. 

46.  Solve  for  a  and  m :  2  a2  —  am  — 12, 

2  am  —  m2  =  8. 

47.  Factor  ,  A  •*"       ■  -  - — ±— —  • 

(V  +  3)-1      (1+3  a)"1 

48.  Rationalize  the  denominator  of  — i L^l — . 

2-V6  +  2V2 

49.  Insert  6  geometrical  means  between  T\  and  12|. 

Ca    -cv   a      •     xy/S      -v/18 

50.  Find  x  in  — ^—  =  ~ 

12  y/i 


GENERAL  REVIEW  401 

51.   Without  solving,  prove  that  the  roots  of  6  x2  +  5  x  —  21 
are  real  and  rational. 


52.  Solve  m-f  \Z3x-\-x2 :  m  +  l=m-  V3  #-}-#* :  m  — 1. 

53.  The  difference  between  the  5th  and  the  7th  terms  of  an 
arithmetical  progression  is  6,  and  the  sum  of  the  first  14  terms 
is  — 105.     Find  the  first  term  and  the  common  difference. 

54.  Without  solving,  determine  the  nature  of  the  roots  of 
16ar>  +  l=8a;. 

55.  Find  the  5th  term  of  (x2-  ar2)15. 

56.  Show  that  the  sum  of  the  squares  of  the  roots  of 
»2-3iC  +  l  =  0  is  7. 

57.  If  m2  —  n2  varies  as  x2,  and  if  x  =  2  when  ra  =  5  and 
ra  =  3,  find  the  equation  between  m,  n,  and  x. 

58.  Find  x  and  y  if  2x+»  =  16  and  Sx~v  =  9. 

59.  Find  a  4th  proportional  to  x*  —  1,  x2  -f  1,  and  cc2  —  1. 

60.  Construct  the  quadratic  equation,  the  product  of  whose 
roots  shall  be  twice  the  sum  of  the  roots  of  x2  —  7  sc  -f- 12  =  0 ; 
and  the  sum  of  whose  roots  shall  be  3  times  the  product  of 
the  roots  of  x2  -f  2  x  =  3. 

61.  Factor  3 x*-2  +  x2  +  x4 -3 x. 

62.  Find  by  logarithms  the  value  of  -y/2  x  (£)*  X  .01  x  3*. 

63.  The  sum  of  two  numbers  is  20,  and  their  geometrical 

mean  increased  by  2  equals  their  arithmetical  mean.     What 

are  the  numbers  ? 

i  _ 

64.  What  is  the  interpretation  of  x1  =  Vsc? 

65.  Write  the  5th  term  of  (a  -f  b)m. 

66.  Solve  c2  O2  + 1)  =  m2  +  2  c2x. 

67.  The  sum  of  the  last  3  terms  of  an  arithmetical  progres- 
sion of  7  terms  equals  3  times  the  sum  of  the  first  3  terms. 
The  sum  of  the  3d  and  5th  terms  is  32.  Find  the  1st  term 
and  the  common  ratio. 

SOM.    EL.    ALG.  —  26 


402  GENERAL  REVIEW 

68.  How  many  digits  in  35s5  ? 

69.  Expand  and  simplify  (a_1Va  —  aVa-1)4. 

70.  What  must  be  the  equation  between  m  and  n  if  the  roots 
of  mxP+nx+p  are  real  ?  if  equal  ? 

71.  Find  the  (r  +  l)th  term  of  (1  -x)20. 

72.  Find  two  numbers  in  the  ratio  of  3:2,  such  that  their 
sum  has  to  the  difference  of  their  squares  the  ratio  of  1 : 5. 

73.  What  is  the  sum  and  product  of  the  roots  of   ,> 

5  x~2  x~l 

74.  Solve  x2  +  y2  =  26, 

5  x  +  y  =  24. 

Plot  the  graphs  of  the  system  and  verify  the  solutions. 

75.  Show  that  either  root  of  x2  —  c  =  0  is  a  mean  propor- 
tional between  the  roots  of  x2  +  bx  +  c  =  0. 


76.  Solve  x2  —  2  x  +  3  ==  V&-2  —  2  a;  +  5.  Are  all  the  solu- 
tions roots  of  the  given  equation  ? 

77.  What  is  the  value  of  the  6th  term  of  f  x ]    when 

x  =  2? 

78.  The  intensity  of  light  varies  inversely  as  the  square  of 
the  distance  from  its  source.  How  far  must  an  object  that  is 
8  feet  from  a  lamp  be  moved  so  that  it  may  receive  but  \  as 
much  light? 

79.  Find  an  expression  for  x  in  a2x  =  3  c. 

80.  Solve  x  — 1.3  =  .3  x~\ 

81.  If  a  :  b  =  c > :  d,  show  that 

(ma  —  nb)(ma  +nb)~l  _  (mc  —  nd)(mc  +  wcT)"1 
mn  win 

10 


82.    Solve  V7  x  -  V3  a  +  4  = 


V3a+4 


GENERAL  REVIEW  403 

83.  If  rx  and  r2  represent  the  roots  of  x2  -f  bx  -f  c  =  0,  find 
r*  -f-  r22  an(i  ^i2**22» 

84.  Solve  x~%  +  af*  + 1  =  0. 

85.  Find  the  geometrical  progression  whose  sum  to  infinity 
is  4^,  and  whose  2d  term  is  .002. 

86.  Expand  (  V2  -  V^)6. 


87.  Solvefors:   -2  -  2  s2+  V7  +2  s  +  4  s2-  2s  =  2s2+  5. 

88.  Find  the  value  of  k  in  order  that  the  equation 

(k  +  6)x2-2k(x2-l)-2kx-3  =  0 
may  have  equal  roots. 

89.  The  floor  area  of  a  certain  room  is  320  sq.  ft.,  each 
end  wall  128  sq.  ft.,  and  each  side  wall  160  sq.  ft.  What 
are  the  dimensions  of  the  room  ? 

90.  For  what  values  of  m  are  the  roots  of  the  equation 
(m  +  2)x2  +  2ma;+l  =  0  equal? 

91.  Given  a:b  =  c: dj  prove  that  3a  +  2c:3a  —  2c 
=  12  b  +  8  d :  12  b  -  8  d. 

92.  Plot  the  graph  of  3  x2  -f- 10  x  =  12,  and  check  the  result 
by  solving. 

93.  Find  the  ratio  between  the  5th  term  of  f  1  +  - )  and 
the  4th  term  of  (l  +  |Y2- 

94.  Calculate  by  logarithms  the  fourth  proportional  to  3.84, 
2.76  and  4.62,  and,  also,  the  mean  proportional  between  -\/T2 
and  V12. 

95.  Solve  for  sand  t:  sP  +  P  =  91,  s  =  7  —  t. 

96.  Findnins  =  arW~a» 


404  GENERAL  REVIEW 

97.  The  velocity  of  a  body  falling  from  rest  varies  directly 
as  the  time  of  falling.  If  the  velocity  of  a  ball  is  160  feet 
after  5  seconds  of  fall,  what  will  it  be  at  the  end  of  the  10th 
second  ? 

98.  In  an  arithmetic  progression,  a—  —  V—  1,  cZ=l+ V^T, 
n  a*  20.     Find  I  and  s. 

99.  Write  the  (r  +  l)th  term  of  (a  +  b)n. 

100.  Find  the  middle  term  of  ( — X~3Y°. 

W-l      ,*    J 

101.  Plot  the  graphs  of  4  x2  +  9  f  =  36,  x  +  2  y  =  3.  Check 
by  solution. 

102.  Insert  4  geometrical  means  between  V— 1  and  —  32. 

103.  Prove  the  formula  for  I  in  each  of  the  progressions. 

104.  Form  the  quadratic  equation  which  has  for  one  root 
the  positive  value  of  V  J  4-  4V3,  and  for  the  other  root  the 
arithmetic  mean  between  4  —  2  V3  and  zero. 

105.  If  m :  n  —  n :  s  =  s  :  t,  show  that  n  +  s  is  a  mean  pro- 
portional between  m  +  n  and  t  -\-  s. 

106.  Solve  xy  =  2  m2  +  5  m  +  2,  ar>  +  ?/2  =  5  ra2  +  8  m  +  5. 


107.  Solve  and  test  the  solution :  V#2  —  mx  -f»  *-  x*s  m, 

f         2\12 

108.  Find  the  term  of  (x4 J     that  does  not  contain  x. 

109.  Solve  for  sand  t:  s2  +  st  +  2t2  =  46, 

2s2-st  +  t2=29. 

110.  Plot  the  graphs  of  a^  +  y2  +  x  +  y  =34,  x  +  y-7  =  0. 

111.  What  is  the  value  of  1.027027  •;•  ? 

112.  Find  x  if  32*~2  j==  (9~ly-\ 

113.  If  a  -f  b  —  61,  and  a^  —  6^  =  1,  find  the  values  of  a 


and  b. 


INDEX 


(Numbers  refer  to  pages.) 


Abscissa,  197. 
Addends,  22. 
Addition,  22. 

Affected  quadratic  equation,  2 
Aggregation,  signs  of,  29. 
Algebraic  expression,  20. 
Algebraic  fraction,  122. 
Algebraic  number,  16. 
Alternation,  320. 
Annuity,  382. 
Antecedent,  315,  318. 
Antilogarithm,  373. 
Arithmetical  means,  344. 
Arithmetical  progression,  338. 
Ascending  order,  44. 
Associative  law,  22,  39. 
Axes  of  reference,  197. 
Axiom,  13,  65. 

Base  of  logarithm,  364.     * 
Binomial,  21. 
Binomial  formula,  359. 
Binomial  theorem,  358. 
Brace,  29. 
Bracket,  29. 

Characteristic  of  logarithm,  3 
Checking  results,  25,  46,  59. 
Clearing  of  fractions,  147. 
Coefficient,  12.- 

compound,  35. 

detached,  51,  61. 
Cologarithm,  376. 
Collecting  terms,  25. 
Common  difference,  338. 
Common  factor,  119. 
Common  multiple,  127. 
Common  ratio,  348. 
Commutative  law,  22,  39. 
Complex  fraction,  144. 
Complex  number,  255. 
Composition,  321. 
Compound  ratio,  348. 
Comoound  variation,  333. 


Conditional  equation,  64. 
Conjugate  imaginary,  260. 
Consequent,  315,  318. 
Constant,  200,  300. 
Continued  proportion,  318. 
Coordinates,  rectilinear,  197. 
Cube  root,  99,  394,  396. 

Definite  number  symbols,  9. 
Degree,  of  an  expression,  44. 

of  a  term,  44. 
Denominator,  126. 

factorial,  360. 

lowest  common,  129. 
Density,  164. 
Descending  order,  44. 
Difference,  32. 

common,  338. 
Direct  and  inverse  variation,  332. 
Direct  variation,  331. 
Discriminant,  278. 
Discussion  of  a  problem,  193. 
Distributive  law,  39,  52. 
Dividend,  52. 
Division,  52,  321. 
Divisor,  52. 

Divisors,  theory  of,  387. 
Duplicate  ratio,  316. 

Elements,  of  an  arithmetical  'progres- 
sion, 339. 

of  a  geometrical  progression,  349. 
Elimination,  165. 
Equality,  12. 
Equation,  64. 

affected  quadratic,  268. 

complete  quadratic,  268. 

conditional,  64. 

equivalent,  174,  290. 

exponential,  380. 

identical,  64. 

in  the  quadratic  form ,  286. 

incomplete  quadratic,  266. 

inconsistent,  174. 
405 


406 


INDEX 


Equation,  independent,  173. 

irrational,  252. 

linear,  65. 

pure  quadratic,  266. 

quadratic,  266. 

simple,  65. 

simultaneous,  174. 

solution  of  an,  64. 

systems  of,  174. 
Equilibrium,  165. 
Equivalent  equation,  174. 
Exponent,  41. 

in  the  fractional  form,  223. 

negative,  222. 

zero,  221. 
Exponential  equation,  380. 
Expression,  algebraic,  20. 

homogeneous,  45. 

integral,  99. 

mixed,  125. 

prime,  100. 

rational,  99. 
Extraneous  roots,  181. 
Extremes,  of  a  proportion,  317. 

Factor,  12,  99. 

common,  119. 

highest  common,  119. 

theorem,  385. 
Formula,  87. 

binomial,  359. 

physical,  164,  293. 

quadratic,  273. 
Fourth  proportional,  318. 
Fourth  root,  99. 
Fraction,  algebraic,  122. 

complex,  144. 

terms  of  a,  122. 
Fractions,  clearing  of,  147. 

General  number  symbols,  9. 
Geometric  means,  354. 
Geometric  progression,  348. 
Graph,  of  a  linear  equation,  200. 

of  a  point,  197. 

of  a  quadratic  equation  in  one  vari- 
able, 282. 

of  a  quadratic  equation  in  two  vari 
ables,  302. 
Grouping,  law  of,  22,  39. 


Identical  equation,  64. 

Identity,  64. 

Imaginaries,  conjugate,  260. 

unit  of,  254. 
Imaginary  number,  254. 

pure,  254. 
Inconsistent  equation,  174. 
Independent  equation,  174. 
Indeterminate  equation,  173. 
Index  law,  42,  53,  205,  210. 
Indicated  operations,  12. 
Integral  expression,  99. 
Interpolation,  369,  373. 
Inverse  ratio,  316. 
Inverse  variation,  332. 
Inversion,  320. 
Involution,  205. 
Irrational  equation,  252. 

Joint  variation,  332. 

Law,  associative,  22,  39. 

commutative,  22,  39. 

distributive,  39,  52. 
Linear  equation,  65. 
Literal  expression,  20. 
Literal  number  symbols,  9. 
Logarithm,  364. 

base  of  a,  364. 

characteristic  of  a,  366. 

common,  365. 

mantissa  of  a,  366. 
Lowest  common  denominator,  129. 
Lowest  common  multiple,  127. 

Mantissa  of  logarithm,  366. 
Mean  proportional,  318. 
Means,  arithmetical,  344. 

geometrical,  354. 

of  a  proportion,  317. 
Mixed  expression,  125. 
Moment,  165. 
Monomial,  21. 
Multiple,  common,  127. 

lowest  common,  127. 
Multiplicand,  38. 
Multiplication,  38. 
Multiplier,  38. 

Negative  exponents,  222. 


INDEX 


407 


Negative  numbers,  14. 
Number,  algebraic,  16. 

negative,  14. 
Numerator,  126. 
Numerical  expression,  20. 

Order,  ascending,  44. 

descending,  44. 

law  of,  22,  38. 
Ordinate,  197. 
Origin,  197. 

Parenthesis,  13,  26. 

Physical  formulas,  164,  293. 

Polynomial,  21. 

Power,  41. 

Prime  expression,  100. 

Principal  root,  237. 

Problem,  70. 

discussion  of  a,  193. 

solution  of  a,  71. 
Progression,  arithmetical,  338. 

geometrical,  348. 
Proportion,  317. 

continued,  318. 

extremes  of  a,  317. 

means  of  a,  317. 
Pure  imaginary,  254. 
Pure  quadratic  equation ,  266. 

Quadrants,  198. 
Quadratic  equation,  266. 
Quadratic  form,  286. 
Quadratic  formula,  273. 
Quality,  signs  of,  14. 
Quotient,  52. 

Radical,  236. 

index  of  a,  209,  236. 

sign  of,  209. 

similar,  242. 
Radicand,  236. 
Ratio,  common,  348. 

compound,  316. 

duplicate,  316. 

inverse,  316. 
Rational  expression,  99. 
Rationalization,  246. 
Real  number,  254. 


Reciprocal,  140. 
Rectilinear  coordinates,  197. 
Remainder,  32. 
Root,  41,  65. 

cube,  99,  394,  396. 

square,  99. 
Roots,  rejection  of,  291. 

Series,  finite,  258. 

infinite  decreasing,  352. 
Signs,  of  aggregation,  29. 

of  quality,  14. 
Similar  terms,  20. 
Simple  equation,  65. 
Simultaneous  equations,  174. 
Solution  of  equations,  64. 
Square  root,  99. 
Substitution,  84. 
Subtraction,  32. 
Subtrahend,  32. 
Sum,  22. 
Surd,  236. 

coefficient  of  a,  236. 

entire,  236. 

mixed,  236. 
Surds,  conjugate,  247. 
Symbols  of  operation,  11. 
Systems  of  equations,  174. 

Term,  20. 

degree  of  a,  44. 
Terms,  of  a  fraction,  122. 

similar,  20. 
Third  proportional,  318. 
Transposition,  66. 
Trinomial,  21. 

Unit  of  imaginaries,  254. 

Variable,  200,  330. 
Variation,  compound,  333. 

direct,  331. 

direct  and  inverse,  332. 

inverse,  332. 

joint,  332. 
Verification  of  a  root,  67. 
Vinculum,  17. 

|  Zero  exponent,  221. 


ROBBINS'S  GEOMETRY 

By  EDWARD  RUTLEDGE  ROBBINS,  A.B.,  Senior 
Mathematical  Master,  The  William  Benn  Charter  School, 
Philadelphia. 


Plane  and  Solid  Geometry,     $1.25        Plane  Geometry     .      .      .      $0.75 
Solid  Geometry      .      .      .     #0.75 


THIS  text-book  is  intended  to  meet  the  needs   of  all 
secondary-  schools  and  the  requirements  in  geometry 
for  entrance  to  all  colleges  and  universities.     It  is  clear, 
consistent,  teachable,  and  sound. 

^j  So  suggestively  and  comprehensively  is  the  work  outlined 
that  the  teacher  is  saved  many  explanations,  while  the  pupil 
receives  the  help  he  desires  wherever  it  is  required.  The 
preliminary  matter  is  intentionally  brief  and  simple,  so  that 
the  theorems  and  their  demonstrations  may  be  reached  as  early 
as  possible.  Each  theorem  is  employed  in  the  demonstration 
of  other  theorems  as  promptly  as  is  practicable  and  desirable, 
^j  The  book  has  been  so  constructed  that  it  meets  the  various 
degrees  of  intellectual  capacity  and  maturity  in  all  ordinary 
classes.  The  successive  truths  in  a  demonstration  are  stated, 
and  the  pupil  is  asked  the  reasons.  But  the  latter  is  not  left 
in  ignorance  if  he  is  unable  to  perceive  the  correct  reason, 
because  the  numbers  of  the  paragraphs  containing  these  desired 
truths  are  cited  wherever  experience  has  shown  that  the 
pupil  is  likely  to  require  this  assistance. 

^j  The  conciseness  and  rigor  of  demonstration,  the  great  wealth 
of  original  exercises,  both  classified  and  graded,  the  economy 
of  arrangement,  the  full  treatment  of  measurement,  and  the 
superior  character  of  the  diagrams  distinguish  the  book.  Pre- 
ceding the  earlier  collections  of  original  exercises  are  summaries, 
while  useful  formulas  are  grouped  both  in  the  Plane  and  in 
the  Solid.  All  the  theorems  in  the  Plane  Geometry  are 
stated  at  the  beginning  of  the  Solid. 


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ELEMENTS      OF 
TRIGONOMETRY 

By  ANDREW  W.   PHILLIPS,  Ph.D.,  and  WENDELL 
M.  STRONG,  Ph.D.,  Professors  in  Yale  University 


Plane  and  Spherical  Trigonometry.    With  Tables $i-4Q 

The  same.     Without  Tables 90 

Logarithmic  and  Trigonometric  Tables 1.00 


IN  this  text-book  full  recognition  is  given  to  the  rigorous 
ideas  of  modern  mathematics  in  dealing  with  the  funda- 
mental series  of  trigonometry.  Both  plane  and  spherical 
trigonometry  are  treated  in  a  simple,  direct  manner,  free  from 
all  needless  details.  The  trigonometric  functions  are  defined 
as  ratios,  but  their  representation  by  lines  is  also  introduced 
at  the  beginning,  because  certain  parts  of  the  subject  can  be 
treated  more  simply  by  the  line  method,  or  by  a  combination 
of  the  two  methods,  than  by  the  ratio  method  alone. 
^J  Many  valuable  features  distinguish  the  work,  but  attention 
is  called  particularly  to  the  graphical  solution  of  spherical 
triangles,  the  natural  treatment  of  the  complex  number,  and 
the  hyperbolic  functions,  the  graphical  representation  of  the 
trigonometric,  inverse  trigonometric,  and  hyperbolic  functions, 
the  emphasis  given  to  the  formulas  essential  to  the  solution  of 
triangles,  the  close  and  rigorous  treatment  of  imaginary  quan- 
tities, the  numerous  cuts  which  simplify  the  subject,  and  the 
rigorous  chapter  on  computation  of  tables, 
•fl"  Carefully  selected  exercises  are  given  at  frequent  intervals, 
affording  adequate  drill  just  where  it  is  most  needed.  An 
exceptionally  large  number  of  miscellaneous  exercises  are 
included  in  a  separate  chapter. 

^[  The  tables  include,  besides  the  ordinary  five-place  tables, 
a  complete  set  of  four-place  tables,  a  table  of  Naperian 
logarithms,  tables  of  the  exponential  and  hyperbolic  functions, 
a  table  of  constants,  etc. 


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ELEMENTS  OF  DESCRIPTIVE 
GEOMETRY 

By    CHARLES    E.    FERRIS,    Professor    of   Mechanical 
Engineering,   University  of  Tennessee 

#1.25 


THE  leading  engineers  and  draughtsmen,  as  investigation 
shows,  do  nearly  all  their  work  in  the  third  quadrant  or 
angle.  It  seems  reasonable,  therefore,  that  the  subject 
of  descriptive  geometry  should  be  taught  in  technical  and 
scientific  schools  as  it  will  be  used  by  their  graduates, 
•[f  Many  years  of  experience  in  teaching  descriptive  geometry 
have  proved  to  the  author  that  the  student  can  learn  to  think 
with  his  problem  below  the  horizontal,  and  behind  the  vertical 
and  perpendicular  planes,  as  well  as  above  and  in  front  of 
those  planes. 

^[  This  volume  forms  an  admirable  presentation  of  the  subject, 
treating  of  definitions  and  first  principles  ;  problems  on  the 
point,  line,  and  plane ;  single  curved  surfaces ;  double  curved 
surfaces  ;  intersection  of  single  and  double  curved  surfaces  by 
planes,  and  the  development  of  surfaces  ;  intersection  of  solids  ; 
warped  surfaces  ;  shades  and  shadows  ;  and  perspective. 
^|  Besides  dealing  with  all  its  problems  in  the  third  angle 
instead  of  in  the  first,  the  book  presents  for  each  problem  a 
typical  problem  with  its  typical  solution,  and  then  gives 
numerous  examples,  both  to  show  variations  in  the  data,  and  to 
secure  adaptability  in  the  student.  In  consequence,  no  sup- 
plementary book  is  necessary. 

*H  To  show  the  projections  on  the  horizontal,  and  on  the 
vertical  planes,  it  uses  v  and  h  as  exponents  or  subscripts 
instead  of  the  usual  method  of  prime,  etc. 
^[  In  scope  the  treatment  is  sufficiently  broad,  and  yet  it  is 
not  so  abstruse  as  to  make  the  book  difficult  for  the  average 
college  course.  Both  text  and  plates  are  bound  together,  thus 
being  very  convenient  for  use.      There  are  1 1  3  figures. 


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PLANE     SURVEYING 

#3.00 

By  WILLIAM  G.  RAYMOND,  C.  E.,  member  Ameri- 
can Society  of  Civil  Engineers,  Professor  of  Geodesy, 
Road  Engineering,  and  Topographical  Drawing  in 
Rensselaer  Polytechnic  Institute. 


IN  this  manual  for  the  study  and  practice  of  surveying  the 
subject  is  presented  in  a  clear  and  thorough  manner;  the 
general  method  is  given  first  and  afterward  the  details. 
Special  points  of  difficulty  have  been  dwelt  on  wherever 
necessary.  The  book  can  be  mastered  by  any  student  who 
has  completed  trigonometry,  two  formulas  only  being  given, 
the  derivation  of  which  requires  a  further  knowledge.  The 
use  of  these  is,  however,  explained  with  sufficient  fullness. 
^|  In  addition  to  the  matter  usual  to  a  full  treatment  of  land, 
topographical,  hydrographical,  and  mine  surveying,  par- 
ticular attention  is  given  to  system  in  office-work,  labor-saving 
devices,  the  planimeter,  slide-rule,  diagrams,  etc.,  coordinate 
methods,  and  the  practical  difficulties  encountered  by  the 
young  surveyor.  An  appendix  gives  a  large  number  of 
original  problems  and  illustrative  examples. 
^J  The  first  part  describes  the  principal  instruments  and  deals 
with  the  elementary  operations  of  surveying,  such  as  measure- 
ment of  lines,  leveling,  determination  of  direction  and  measure- 
ment of  angles,  stadia  measurements,  methods  of  computing 
land  surveys,  etc. 

•[f  In  the  second  part  are  treated  general  surveying  methods, 
including  land  surveys,  methods  adapted  to  'farm  surveys, 
United  States  public  land  surveys,  and  city  surveys,  curves, 
topographical  surveying,  ordinary  earthwork  computations, 
hydrographic  and  mine  surveying,  etc. 

^[  Both  four-place  and  five-place  tables  are  provided.  They 
are  unusually  numerous  and  practical,  and  are  set  in  large, 
clear  type.      The  illustrations  are  particularly  helpful. 


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ESSENTIALS    IN    HISTORY 


ESSENTIALS  IN  ANCIENT  HISTORY  .     .     $i.;o 

From  the  earliest  records  to  Charlemagne.  By  ARTHUR 
MAYER  WOLFSON,  Ph.D.,  First  Assistant  in  History, 
DeWitt  Clinton  High  School,  New  York. 

ESSENTIALS  IN   MEDIEVAL  AND   MODERN 
HISTORY $1.50 

From  Charlemagne  to  the  present  day.  By  SAMUEL 
BANNISTER  HARDING,  Ph.D.,  Professor  of  Euro- 
pean History,  Indiana  University. 

ESSENTIALS  IN  ENGLISH  HISTORY    .     .     fi.50 

From  the  earliest  records  to  the  present  day.  By 
ALBERT  PERRY  WALKER,  A.M.,  Master  in  His- 
tory, English  High  School,  Boston. 

ESSENTIALS   IN   AMERICAN    HISTORY  .     $1.50 

From  the  discovery  to  the  present  day.  By  ALBERT 
BUSHNELL  HART,  LL.D.,  Professor  of  History, 
Harvard  University. 


THESE  volumes  correspond  to  the  four  subdivisions  re- 
quired   by  the   College  Entrance  Examination   Board, 
and  by  the  New  York  State  Education  Department. 
Each  volume  is  designed  for  one  year's  work.      Each  of  the 
writers  is  a  trained  historical  scholar,  familiar  with  the  con- 
ditions and  needs  of  secondary  schools. 

^[  The  effort  has  been  to  deal  only  with  the  things  which 
are  typical  and  characteristic;  to  avoid  names  and  details 
which  have  small  significance,  in  order  to  deal  more  justly 
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kind. Especial  attention  is  paid  to  social  history,  as  well  as 
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^|  The  books  are  readable  and  teachable,  and  furnish  brief 
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a  significant  and  thorough  body  of  illustration,  which  shall 
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A    BRIEF    COURSE    IN 
GENERAL    PHYSICS 

$1.20 

By    GEORGE    A.    HOADLEY,    A.M.,    C.E., 
Professor  of  Physics,  Swarthmore  College 


A  COURSE,  containing  a  reasonable  amount  of  work  for 
an  academic  year,  and  covering  the  entrance  require- 
ments of  all  of  the  colleges.  It  is  made  up  of  a  reliable 
text,  class  demonstrations  of  stated  laws,  practical  questions 
and  problems  on  the  application  of  these  laws,  and  laboratory 
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^J  The  text,  which  is  accurate  and  systematically  arranged, 
presents  the  essential  facts  and  phenomena  of  physics  clearly 
and  concisely.  While  no  division  receives  undue  prominence, 
stress  is  laid  on  the  mechanical  principles  which  underlie  the 
whole,  the  curve,  electrical  measurements,  induced  currents, 
the  dynamo,  and  commercial  applications  of  electricity. 
•J]"  The  illustrative  experiments  and  laboratory  work,  intro- 
duced at  intervals  throughout  the  text,  are  unusually  numerous, 
and  can  be  performed  with  comparatively  simple  apparatus. 
Additional  laboratory  work  is  included  in  the  appendix,  to- 
gether with  formulas  and  tables. 


HOADLEY'S    PRACTICAL    MEASUREMENTS    IN 
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THIS  book,  which  treats  of  the  fundamental  measurements  in  elec- 
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and  manual  training  schools,  as  well  as  for  colleges.  Nearly  ioo  experiments 
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followed  by  a  simple  discussion  of  the  principles  involved,  and,  in  some 
cases,  by  a  statement  of  well-known  results. 


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CHEMISTRIES 

By  F.  W.  CLARKE,  Chief  Chemist  of  the  United  States 
Geological  Survey,  and  L.  M.  DENNIS,  Professor  of 
Inorganic  and  Analytical  Chemistry,  Cornell  University 


Elementary  Chemistry     .  $1.10 


Laboratory  Manual     .     .  $0.50 


THESE  two  books  are  designed  to  form  a  course  in 
chemistry  which  is  sufficient  for  the  needs  of  secondary 
schools.  The  TEXT- BO  OK  is  divided  into  two  parts, 
devoted  respectively  to  inorganic  and  organic  chemistry. 
Diagrams  and  figures  are  scattered  at  intervals  throughout  the 
text  in  illustration  and  explanation  of  some  particular  experi- 
ment or  principle.  The  appendix  contains  tables  of  metric 
measures  with  English  equivalents. 

^|  Theory  and  practice,  thought  and  application,  are  logically 
kept  together,  and  each  generalization  is  made  to  follow  the 
evidence  upon  which  it  rests.  The  application  of  the  science 
to  human  affairs,  its  utility  in  modern  life,  is  also  given  its 
proper  place.  A  reasonable  number  of  experiments  are  in- 
cluded for  the  use  of  teachers  by  whom  an  organized  laboratory 
is  unobtainable.  Nearly  all  of  these  experiments  are  of  the 
simplest  character,  and  can  be  performed  with  home-made 
apparatus. 

^f  The  LABORATORY  MANUAL  contains  127  experi- 
ments, among  which  are  a  few  of  a  quantitative  character.  Full 
consideration  has  been  given  to  the  entrance  requirements  of 
the  various  colleges.  The  left  hand  pages  contain  the  experi- 
ments, while  the  right  hand  pages  are  left  blank,  to  include 
the  notes  taken  by  the  student  in  his  work.  In  order  to  aid 
and  stimulate  the  development  of  the  pupil's  powers  of  observa- 
tion, questions  have  been  introduced  under  each  experiment. 
The  directions  for  making  and  handling  the  apparatus,  and 
for  performing  the  experiments,  are  simple  and  clear,  and  are 
illustrated  by  diagrams  accurately  drawn  to  scale. 


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